If there is one prayer that you should pray/sing every day and every hour, it is the LORD's prayer (Our FATHER in Heaven prayer)
It is the most powerful prayer. A pure heart, a clean mind, and a clear conscience is necessary for it.
- Samuel Dominic Chukwuemeka

For in GOD we live, and move, and have our being. - Acts 17:28

The Joy of a Teacher is the Success of his Students. - Samuel Dominic Chukwuemeka

# Solved Examples on Probability

For ACT Students
The ACT is a timed exam...60 questions for 60 minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for any wrong answer.

For JAMB Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE-FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions.
Show all work.
Unless specified or implied otherwise, leave all applicable answers as simplified fractions or integers.

NOTE:
(1.) For all the questions:
S is the sample space
n(S) is the cardinality of the sample space.

(2.) In some cases, it is recommended that you do not simplify until the final answer.

(3.) Prerequisites:
(a.) Fractions, Decimals, and Percents
(b.) Expressions and Equations
(c.) Set Algebra for some questions.

(4.) Corequisites:
(a.) Combinatorics
Some of these questions are preferably solved using Combinatorics.
You can review those same questions solved using Combinatorics
(b.) Binomial Distribution Applications
Some of these questions may be solved using Binomial Probability Distribution.

(1.) Probability in Biology People who inherit one sickle cell gene and one normal gene have sickle cell trait (SCT).
People with SCT usually do not have any of the symptoms of sickle cell disease (SCD), but they can pass the trait on to their children.
[What is Sickle Cell Trait? (https://www.cdc.gov/ncbddd/sicklecell/traits.html)]
Sickle cell disease (SCD) is a group of inherited red blood cell disorders.
Red blood cells contain hemoglobin, a protein that carries oxygen.
Healthy red blood cells are round, and they move through small blood vessels to carry oxygen to all parts of the body.
In someone who has SCD, the hemoglobin is abnormal, which causes the red blood cells to become hard and sticky and look like a C-shaped farm tool called a sickle
The sickle cells die early, which causes a constant shortage of red blood cells.
Also, when they travel through small blood vessels, they get stuck and clog the blood flow. This can cause pain and other serious complications (health problems) such as infection, acute chest syndrome and stroke.
[What is Sickle Cell Disease? (https://www.cdc.gov/ncbddd/sicklecell/facts.html)]

A man and a woman wants to marry.
Both have the sickle cell trait (SCT).
They do not know about Probability in Biology, but they saw this diagram at the CDC’s website.
They know you took a Statistics class with Mr. C and that you might help explain the diagram.

(a.) Using a Tree Diagram and/or a Punnett Square, explain the diagram to the man and his fiancée.
Include the concept of Probability in your explanations. Assume they intend to have four children.

(b.) Should they get married or not? Advise them.

Genotype of Man = AS
The man has a sickle cell trait, SCT (because of the S)
Genotype of Woman = AS
The woman also has a sickle cell trait, SCT (because of the S)
Assume they intend to have four children
Let us draw a Punnett Square to illustrate the likely genotypes of their four children

Woman →
Man ↓
$A$ $S$
$A$ $AA$ $AS$
$S$ $AS$ $SS$

If they have four children, it is likely (not certain) that:
1 child will have genotype AA
This means that one child is likely to be typical (no blood disorder).

2 children will have genotype AS
This means that two children are likely to inherit the sickle cell traits.

1 child will have genotype SS
This means that one child is likely to have the sickle cell disease.
The issue is with the SS genotype. The child with that genotype will likely need a lot of medical care.
Why bring a child into this world to suffer the child?
Do not recommend the marriage.

$n(SS) = 1 \\[3ex] n(S) = 4 \\[3ex] P(SS) = \dfrac{n(SS)}{n(S)} \\[5ex] P(SS) = \dfrac{1}{4} = 0.25 = 25\% \\[3ex]$ If they have four children, there is a 25% probability that one of the children will have the sickle cell disease.
(2.) Probability in Biology A man with $AS$ genotype wants to marry a woman with $AS$ genotype.
They intend to have at least four children if they marry.
(a.) If they have four children, what are the likely genotypes of each of their children?
(b.) Discuss the probability of getting a child with a sickle cell trait.
(c.) Would you recommend the marriage?
(d.) Draw a tree diagram or a Punnett Square to discuss your answers.
You may review the document by S.I.R. Okoduwa

Genotype of Man = $AS$
Genotype of Woman = $AS$
Let us draw a Punnett Square to illustrate the likely genotypes of their four children

Woman →
Man ↓
$A$ $S$
$A$ $AA$ $AS$
$S$ $AS$ $SS$

If they have four children, it is likely (not certain) that:
1 child will have genotype AA
2 children will have genotype AS
1 child will have genotype SS

The issue is with the SS genotype. The child with that genotype will likely need a lot of medical care.
Why bring a child into this world to suffer the child?
Do not recommend the marriage.

$n(SS) = 1 \\[3ex] n(S) = 4 \\[3ex] P(SS) = \dfrac{n(SS)}{n(S)} \\[5ex] P(SS) = \dfrac{1}{4} = 0.25 = 25\% \\[3ex]$ If they have four children, there is a 25% probability that one of the children will have the sickle cell disease.
(3.) A 12-sided die is rolled one time.
Determine the probability of rolling a number less than $10$

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex] n(S) = 12 \\[3ex] E = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] n(E) = 9 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{9}{12} \\[5ex] P(E) = \dfrac{3}{4}$
(4.) A 12-sided die is rolled one time.
Determine the probability of rolling a number greater than $12$

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\} \\[3ex] n(S) = 12 \\[3ex] E = \{ \} \\[3ex] n(E) = 0 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{0}{12} \\[5ex] P(E) = 0 \\[3ex]$ This is an impossible event.
It is impossible to obtain a number greater than $12$ when a $12-sided$ doe is rolled one time.
(5.) Criminal cases are assigned to judges randomly.
The list below is a list of nine judges.
Joshua Judges
Ruth Kings
Samuel Daniel
Nehemiah Job
Esther Proverbs
Isaiah Psalms
Jeremiah Amos
Ezekiel Joel
Micah Nahum

(a.) Event A is the event that the judge is a woman.
List the outcomes of event A

(b.) What is the probability that a case will be assigned to a female judge?

(c.) List the outcomes of the complement of event A

$(a.) \\[3ex] A = \{Ruth\:\:Kings,\:\:Esther\:\:Proverbs\} \\[3ex] (b.) \\[3ex] n(A) = 2 \\[3ex] n(S) = 9 \\[3ex] P(A) = \dfrac{n(A)}{n(S)} \\[5ex] P(A) = \dfrac{2}{9} \\[5ex] (c.) \\[3ex] A' = \{Joshua\:\:Judges,\:\:Samuel\:\:Daniel,\:\:Nehemiah\:\:Job,\:\:Isaiah\:\:Psalms,\:\:Jeremiah\:\:Amos,\:\:Ezekiel\:\:Joel,\:\:Micah\:\:Nahum\}$
(6.) If a coin is flipped 10 times, what percentage of the time will the coin land on heads?
A first step to answering this question is to simulate 10 flips.
Use the Random Number Table below to simulate flipping a coin 10 times.
Let the digits 0, 1, 2, 3, 4 represent heads and the digits 5, 6, 7, 8, 9 represent tails.
Begin with the first digit in the third row of the random number table.

 21033 17516 39329 45506 32522 69328 17705 44716 19305 88389 70251 02498 90633 19770 84413 15327 80873 33197 30896 79149 19167 27336 21490 28409

(a.) Write the sequence of 10 random digits.

(b.) Change the sequence of 10 random digits to a sequence of heads and tails, writing H for the digits 0, 1, 2, 3, 4 and the T for the digits 5, 6, 7, 8, 9.

(d.) What percentage of the flips were heads?

(a.) The sequence of digits is: 3, 9, 3, 2, 9, 1, 7, 7, 0, 5

(b.) The sequence of letters is: H, T, H, H, T, H, T, T, H, T

(c.) The longest streak of heads in the list is 2 heads (3rd and fourth letters)

(d.) The percentage of flips that were heads:

$number\;\;of\;\;flips = 10 \\[3ex] number\;\;of\;\;heads = 5 \\[3ex] \%\;\;of\;\;flips\;\;that\;\;were\;\;heads \\[3ex] = \dfrac{number\;\;of\;\;heads}{number\;\;of\;\;flips} * 100 \\[5ex] = \dfrac{5}{10} * 100 \\[5ex] = 50\%$
(7.) Deborah was carrying out a randomized experiment to test if there is a difference in the amount of information remembered between students who take notes using a computer versus who take notes by hand using pen and paper.
There are 20 college participants and she wants each to have an equal chance of being assigned to the computer group or the pen and paper group.
Let the even digits (0, 2, 4, 6, 8) represent assignment to the computer group and the odd digits (1, 3, 5, 7, 9) represent assignment to the pen and paper group.
Using the Random Number Table, begin with the first digit in the third row.

 21033 17516 26613 45506 32522 69328 04566 44716 19305 88389 24986 02498 90633 19770 78498 15327 80873 33197 30896 79149 19167 27336 21490 28409

(a.) Write the sequence of 20 random digits.

(b.) Change the sequence of 20 random digits to a sequence of C for computer (digits 0, 2, 4, 6, 8) and P for paper (digits 1, 3, 5, 7, 9)

(c.) What percentage of the 20 participants were assigned to the computer group?

(d.) Describe other ways the Random Number Table could have been used to randomly assign participants to one of the two groups.

(a.) The sequence of digits is: 2, 6, 6, 1, 3, 0, 4, 5, 6, 6, 2, 4, 9, 8, 6, 7, 8, 4, 9, 8

(b.) The sequence of letters is: C, C, C, P, P, C, C, P, C, C, C, C, P, C, C, P, C, C, P, C

(c.) The percentage of the 20 participants that were assigned to the computer group:

$number\;\;of\;\;participants = 20 \\[3ex] number\;\;in\;\;computer\;\;group = 14 \\[3ex] \%\;\;of\;\;participants\;\;that\;\;were\;\;computer\;\;group \\[3ex] = \dfrac{number\;\;of\;\;computer\;\;group}{number\;\;of\;\;participants} * 100 \\[5ex] = \dfrac{14}{20} * 100 \\[5ex] = 70\% \\[3ex]$ (d.) Other ways the Random Number Table could have been used to randomly assign participants to one of the two groups include:

(i) The odd dogits could have assigned participants to the computer group, and the even diigits could have assigned participants to the pen and paper group.

(ii) The digits: 0, 1, 2, 3, 4 could have assigned participants to the pen and paper group, and the digits: 5, 6, 7, 8, 9 could have assigned participants to the computer group.

(iii) The digits: 0, 1, 2, 3, 4 could have assigned participants to the computer group, and the digits: 5, 6, 7, 8, 9 could have assigned participants to the pen and paper group.
(8.) ACT A bag contains 16 red marbles, 7 yellow marbles, and 19 green marbles.
How many additional red marbles must be added to the 42 marbles already in the bag so that the probability of randomly drawing a red marble is $\dfrac{3}{5}$?

We can solve this question in two ways
Use any method you prefer

Let the number of additional red marbles = $r$

$\underline{First\:\:Approach} \\[3ex] \underline{Initial} \\[3ex] n(S) = 42 \\[3ex] n(Red) = 16 \\[3ex] \underline{Updated} \\[3ex] n(Red) = 16 + r \\[3ex] n(S) = 42 + r \\[3ex] P(Red) = \dfrac{16 + r}{42 + r} \\[5ex] P(Red) = \dfrac{3}{5} \\[5ex] \therefore \dfrac{16 + r}{42 + r} = \dfrac{3}{5} \\[5ex] Cross\:\: Multiply\:\: method \\[3ex] 5(16 + r) = 3(42 + r) \\[3ex] 80 + 5r = 126 + 3r \\[3ex] 5r - 3r = 126 - 80 \\[3ex] 2r = 46 \\[3ex] r = \dfrac{46}{2} \\[5ex] r = 23 \\[5ex] \underline{Second\:\:Approach} \\[3ex] P(Red) = \dfrac{3}{5} \\[5ex] P(Red) + P(Red') = 1 ...Complementary\:\:Rule \\[3ex] P(Red') = 1 - P(Red) \\[3ex] P(Red') = 1 - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5}{5} - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5 - 3}{5} \\[5ex] P(Red') = \dfrac{2}{5} \\[5ex] n(Red') = n(Yellow) + n(Green) \\[3ex] n(Red') = 7 + 19 = 26 \\[3ex] P(Red') = \dfrac{n(Red')}{n(S)} \\[5ex] n(S) * P(Red') = n(Red') \\[3ex] n(S) = \dfrac{n(Red')}{P(Red')} \\[5ex] n(S) = n(Red') \div P(Red') \\[3ex] n(S) = 26 \div \dfrac{2}{5} \\[5ex] n(S) = 26 * \dfrac{5}{2} \\[5ex] n(S) = 13(5) \\[3ex] n(S) = 65 \\[3ex] Additional\:\:red\:\:balls = 65 - (16 + 7 + 19) \\[3ex] Additional\:\:red\:\:balls = 65 - 42 \\[3ex] Additional\:\:red\:\:balls = 23 \\[3ex]$ $23$ red marbles need to be added so that the probability of drawing a red marble is $\dfrac{3}{5}$
(9.) ACT A professional baseball team will play 1 game Saturday and 1 game Sunday.
A sportswriter estimates the team has a 60% chance of winning on Saturday but only a 35% chance of winning on Sunday.
Using the sportswriter's estimates, what is the probability that the team will lose both games?
(Note: Neither game can result in a tie.)

$A.\:\: 14\% \\[3ex] B.\:\: 21\% \\[3ex] C.\:\: 25\% \\[3ex] D.\:\: 26\% \\[3ex] E.\:\: 39\% \\[3ex]$

Because the options/answers are in percentages, we shall just work with percents

$P(winning) + P(losing) = 100\% ...Complementary\:\: Rule \\[3ex] \underline{Saturday} \\[3ex] P(winning) = 60\% \\[3ex] P(losing) = 100\% - 60\% = 40\% \\[3ex] \underline{Sunday} \\[3ex] P(winning) = 35\% \\[3ex] P(losing) = 100\% - 35\% = 65\% \\[3ex] P(losing\:\: both\:\: games) = 40\% * 65\% ...Multiplication\:\: Rule \\[3ex] = \dfrac{40}{100} * \dfrac{65}{100} \\[5ex] = \dfrac{40 * 65}{100 * 100} \\[5ex] = \dfrac{2600}{1000} \\[5ex] = 0.26 \\[3ex] = 26\%$
(10.) WASSCE A number is selected at random from each of the sets {2, 3, 4} and {1, 3, 5}.
Find the probability that the sum of the two numbers is greater than 3 and less than 7

It is much better to represent this information in a table.
Because the question asked for the probability of the "sum", we shall be adding the elements.

 $(+)$ $2$ $3$ $4$ $1$ $3$ $\color{red}{4}$ $\color{red}{5}$ $3$ $\color{red}{5}$ $\color{red}{6}$ $7$ $5$ $7$ $8$ $9$

Numbers in red are greater than $3$ and less than $7$
Let $x$ be those numbers

$n(S) = 9 \\[3ex] n(3 \lt x \lt 7) = 4 \\[3ex] P(3 \lt x \lt 7) = \dfrac{n(3 \lt x \lt 7)}{n(S)} \\[5ex] P(3 \lt x \lt 7) = \dfrac{4}{9}$
(11.) ACT A bag contains 10 solid-colored marbles of the same size: 3 red, 2 green, 1 yellow, and 4 blue.
Which of the following expressions gives the probability of drawing, at random and without replacement, a blue marble on the 1st draw, a green marble on the 2nd draw, and a blue marble on the 3rd draw?

$A.\:\: \left(\dfrac{4}{10}\right)\left(\dfrac{2}{10}\right)\left(\dfrac{3}{10}\right) \\[5ex] B.\:\: \left(\dfrac{4}{10}\right)\left(\dfrac{2}{10}\right)\left(\dfrac{4}{10}\right) \\[5ex] C.\:\: \left(\dfrac{4}{10}\right)\left(\dfrac{2}{9}\right)\left(\dfrac{4}{8}\right) \\[5ex] D.\:\: \left(\dfrac{4}{10}\right)\left(\dfrac{2}{9}\right)\left(\dfrac{3}{8}\right) \\[5ex] E.\:\: \left(\dfrac{4}{10}\right)\left(\dfrac{3}{9}\right)\left(\dfrac{3}{8}\right) \\[5ex]$

$Let: \\[3ex] red = R \\[3ex] green = G \\[3ex] yellow = Y \\[3ex] blue = B \\[3ex] n(R) = 3 \\[3ex] n(G) = 2 \\[3ex] n(Y) = 1 \\[3ex] n(B) = 4 \\[3ex] n(S) = 3 + 2 + 1 + 4 = 10 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] \underline{Without\:\:Replacement\:\:condition} \\[3ex] blue-green-blue...in\:\:that\:\:order \\[3ex] P(BGB) = \dfrac{4}{10} * \dfrac{2}{9} * \dfrac{3}{8} ...Option\:D \\[5ex] P(BGB) = \dfrac{1 * 1 * 1}{10 * 3 * 1} \\[5ex] P(BGB) = \dfrac{1}{30}$
(12.) Samuel, Dominic, and Chukwuemeka work for a university.
The university wants to send two employees to a Statistics conference.
To be fair, the university decided that the two individuals who would attend will have their names drawn from a hat. This is similar to obtaining a sample random sample of size, 2.
(a.) Determine the sample space of the experiment. In other words, list all the possible simple random samples of size, n = 2.
(b.) What is the probability that Dominic and Chukwuemeka attend the conference?
(c.) What is the probability that Chukwuemeka stays home?

$Let\:\:Samuel = A \\[3ex] Dominic = D \\[3ex] Chukwuemeka = C \\[3ex] (a.)\:\:S = \{AD, AC, DC\} \\[3ex] n(S) = 3 \\[3ex] (b.)\:\:n(DC) = 1 \\[3ex] P(DC) = \dfrac{n(DC)}{n(S)} = \dfrac{1}{3} \\[5ex]$ If Chukwuemeka stays home, this means that Samuel and Dominic will attend

$n(AD) = 1 \\[3ex] (c.)\:\:P(AD) = \dfrac{n(AD)}{n(S)} = \dfrac{1}{3}$
(13.) ACT The 16-member math club needs to choose a student government representative.
They decide that the representative, who will be chosen at random, CANNOT be any of the 3 officers of the club.
What is the probability that Adrian, who is a member of the club but NOT an officer, will be chosen?

$Let\:\: Member = M \\[3ex] Let\:\:Officer = F \\[3ex] n(S) = 16 \\[3ex] n(M) = 13 \\[3ex] n(F) = 3 \\[3ex] n(Adrian) = 1 \\[3ex] Adrian\:\: \epsilon \:\:M \\[3ex] P(Adrian) = \dfrac{n(Adrian)}{n(M)} \\[5ex] P(Adrian) = \dfrac{1}{13d}$
(14.) ACT A bag contains 10 pieces of flavored candy: 4 lemon, 3 strawberry, 2 grape, and 1 cherry.
One piece of candy will be randomly picked from the bag.
What is the probability the candy picked is NOT grape flavored?

$A.\;\; \dfrac{1}{5} \\[5ex] B.\;\; \dfrac{1}{4} \\[5ex] C.\;\; \dfrac{1}{2} \\[5ex] D.\;\; \dfrac{3}{4} \\[5ex] E.\;\; \dfrac{4}{5} \\[5ex]$

$Let\:\: grape = G \\[3ex] n(G) = 2 \\[3ex] n(S) = 10 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} = \dfrac{2}{10} = \dfrac{1}{5} \\[5ex] P(G') = 1 - P(G) \\[3ex] ... Complementary\;\;Rule...Complementary\;\;Events \\[3ex] P(G') = 1 - \dfrac{1}{5} \\[5ex] = \dfrac{5}{5} - \dfrac{1}{5} \\[5ex] = \dfrac{5 - 1}{5} \\[5ex] = \dfrac{4}{5}$
(15.) ACT A bag contains 4 red jelly beans, 5 green jelly beans, and 3 white jelly beans.
If a jelly bean is selected at random from the bag, what is the probability that the jelly bean selected is green?

$F.\:\: \dfrac{1}{12} \\[5ex] G.\:\: \dfrac{1}{5} \\[5ex] H.\:\: \dfrac{5}{23} \\[5ex] J.\:\: \dfrac{5}{12} \\[5ex] K.\:\: \dfrac{5}{7} \\[5ex]$

$Let \\[3ex] red\:\:jelly\:\:bean = R \\[3ex] green\:\:jelly\:\:bean = G \\[3ex] white\:\:jelly\:\:bean = W \\[3ex] S = \{4R, 5G, 3W\} \\[3ex] n(S) = 4 + 5 + 3 = 12 \\[3ex] n(G) = 5 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} \\[5ex] P(G) = \dfrac{5}{12}$
(16.) ACT The 13-member math club needs to choose a student government representative.
They decide that the representative, who will be chosen at random, CANNOT be any of the 3 officers of the club.
What is the probability that Samara, who is a member of the club but NOT an officer, will be chosen?
Before you solve this question, check out Question (13.)
Did you notice anything?

$Let\:\: Member = M \\[3ex] Let\:\:Officer = F \\[3ex] n(S) = 13 \\[3ex] n(M) = 10 \\[3ex] n(F) = 3 \\[3ex] n(Samara) = 1 \\[3ex] Samara\:\: \epsilon \:\:M \\[3ex] P(Samara) = \dfrac{n(Samara)}{n(M)} \\[5ex] P(Samara) = \dfrac{1}{10}$
(17.) WASSCE
 Scores $1$ $2$ $3$ $4$ $5$ $6$ Frequency $25$ $30$ $x$ $28$ $40$ $32$

The table shows the outcome when a die is thrown a number of times.
If the probability of obtaining a 3 is 0.225:
(a.) how many times was the die thrown?
(b.) calculate the probability that a trial chosen at random gives a score of an even number or a prime number.

The number of times the die was thrown is the total frequency
The total frequency is the cardinality of the sample space

$n(S) = 25 + 30 + x + 28 + 40 + 32 \\[3ex] n(S) = 155 + x \\[3ex] P(3) = 0.225 \\[3ex] P(3) = \dfrac{x}{155 + x} \\[5ex] \rightarrow \dfrac{x}{155 + x} = 0.225 \\[5ex] x = 0.225(155 + x) \\[3ex] x = 34.875 + 0.225x \\[3ex] x - 0.225x = 34.875 \\[3ex] 0.775x = 34.875 \\[3ex] x = \dfrac{34.875}{0.775} \\[5ex] x = 45 \\[3ex] n(S) = 155 + x \\[3ex] n(S) = 155 + 45 \\[3ex] (a.)\:\: n(S) = 200 \\[3ex]$ Let the event of getting an even number be $E$
Let the event of getting a prime number be $P$

$E = \{2, 4, 6\} \\[3ex] n(E) = 3 \\[3ex] P = \{2, 3, 5\} \\[3ex] n(P) = 3 \\[3ex] E\:\:\:AND\:\:\:P = \{2\} \\[3ex] n(E\:\:\:AND\:\:\:P) = 1 \\[3ex] P(E\:\:\:OR\:\:\:P) = P(E) + P(P) - P(E\:\:\:AND\:\:\:P) ...Addition\:\:Rule \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{3}{45} \\[5ex] P(P) = \dfrac{n(P)}{n(S)} = \dfrac{3}{45} \\[5ex] P(E\:\:\:AND\:\:\:P) = \dfrac{n(E\:\:\:AND\:\:\:P)}{n(S)} = \dfrac{1}{45} \\[5ex] \rightarrow P(E\:\:\:OR\:\:\:P) = \dfrac{3}{45} + \dfrac{3}{45} - \dfrac{1}{45} \\[5ex] P(E\:\:\:OR\:\:\:P) = \dfrac{3 + 3 - 1}{45} = \dfrac{5}{45} = \dfrac{1}{9} \\[5ex] (b.)\:\: P(E\:\:\:OR\:\:\:P) = \dfrac{1}{9}$
(18.) ACT Mr. Chiang announced the grade distribution for this week's book reports.
Of the 24 students in the class, 8 received A's for their book reports, 11 received B's, and 5 received C's.
When a student is chosen at random to be the first one to read his or her book report to the class, what is the probability that the student chosen had received an A for the book report?

$A.\:\: \dfrac{5}{24} \\[5ex] B.\:\: \dfrac{1}{3} \\[5ex] C.\:\: \dfrac{11}{24} \\[5ex] D.\:\: \dfrac{1}{2} \\[5ex] E.\:\: \dfrac{8}{11} \\[5ex]$

$n(S) = 24 \\[3ex] n(A's) = 8 \\[3ex] P(A's) = \dfrac{n(A's)}{n(S)} = \dfrac{8}{24} = \dfrac{1}{3}$
(19.) The table below shows the Math Guys and the Art Guys among high school students (10th, 11th, and 12th grade students) in a certain school in the City of Truth or Consequences, New Mexico

 10th Grade 11th Grade 12th Grade Total Math Guys $232$ $442$ $59$ $733$ Art Guys $1036$ $68$ $53$ $1157$ Total $1268$ $510$ $112$ $1890$

(a.) If a student is selected at random, what is the probability that the student is a Math Guy?
(b.) If a student is selected at random, what is the probability that the student is in $11^{th}$ grade?
(c.) If a student is selected at random, find the probability that the student is in $11^{th}$ grade OR $12^{th}$ grade.
(d.) If a student is selected at random, determine the probability that the student is in $11^{th}$ grade AND is a Math Guy.
(e.) If a student is selected at random, find the probability that the student is in $11^{th}$ grade OR is a Math Guy.
(f.) If an $11^{th}$ grade student is selected at random, what is the probability that the student is a Math Guy?
(g.) If a $12^{th}$ grade student is selected at random, what is the probability that the student is a Math Guy?
(h.) If a $10^{th}$ grade student is selected at random, determine the probability that the student is a Math Guy
(i.) If two $11^{th}$ grade students are selected at random, determine the probability that both students are Math Guys
(j.) If two $10^{th}$ grade students are selected at random, determine the probability that both students are Math Guys
(k.) If a student is selected at random, determine the probability that the student is in $12^{th}$ grade OR is a Math Guy.

$n(S) = 1890 \\[3ex] n(Math) = 232 + 442 + 59 = 733 \\[3ex] n(Art) = 1036 + 68 + 53 = 1157 \\[3ex] n(10^{th}) = 232 + 1036 = 1268 \\[3ex] n(11^{th}) = 442 + 68 = 510 \\[3ex] n(12^{th}) = 59 + 53 = 112 \\[3ex] (a.) \\[3ex] P(Math) = \dfrac{n(Math)}{n(S)} \\[5ex] P(Math) = \dfrac{733}{1890} \\[5ex] (b.) \\[3ex] n(11^{th}) = 510 \\[3ex] P(11^{th}) = \dfrac{n(11^{th})}{n(S)} \\[5ex] = \dfrac{510}{1890} \\[5ex] = \dfrac{17}{63} \\[5ex] (c.) \\[3ex] n(12^{th}) = 112 \\[3ex] P(12^{th}) = \dfrac{n(12^{th})}{n(S)} \\[5ex] = \dfrac{112}{1890} \\[5ex] n(11^{th}\:\:AND\:\:12^{th}) = 0 \\[3ex] P(11^{th}\:\:\:OR\:\:\:12^{th}) = P(11^{th}) + P(12^{th}) - P(11^{th}\:\:\:AND\:\:\:12^{th}) ... Addition\:\:Rule \\[3ex] P(11^{th}\:\:\:OR\:\:\:12^{th}) = \dfrac{510}{1890} + \dfrac{112}{1890} - 0 \\[5ex] = \dfrac{510 + 112}{1890} \\[5ex] = \dfrac{622}{1890} \\[5ex] = \dfrac{311}{945} \\[5ex] (d.) \\[3ex] n(11^{th}\:\:\:AND\:\:\:Math) = 442 \\[3ex] P(11^{th}\:\:\:AND\:\:\:Math) = \dfrac{n(11^{th}\:\:\:AND\:\:\:Math)}{n(S)} \\[5ex] = \dfrac{442}{1890} \\[5ex] = \dfrac{221}{945} \\[5ex] (e.) \\[3ex] P(11^{th}\:\:\:OR\:\:\:Math) = P(11^{th}) + P(Math) - P(11^{th}\:\:\:AND\:\:\:Math) ... Addition\:\:Rule \\[3ex] P(11^{th}\:\:\:OR\:\:\:Math) = \dfrac{510}{1890} + \dfrac{733}{1890} - \dfrac{442}{1890} \\[5ex] = \dfrac{510 + 733 - 442}{1890} \\[5ex] = \dfrac{267}{630} \\[5ex] = \dfrac{89}{210} \\[5ex] (f.) \\[3ex] P(Math\:\:from\:\:11^{th}) = \dfrac{n(11^{th}\:\:\:AND\:\:\:Math)}{n(11^{th})} \\[5ex] = \dfrac{442}{510} \\[5ex] = \dfrac{13}{15} \\[5ex] (g.) \\[3ex] n(12^{th}\:\:\:AND\:\:\:Math) = 59 \\[3ex] P(Math\:\:from\:\:12^{th}) = \dfrac{n(12^{th}\:\:\:AND\:\:\:Math)}{n(12^{th})} \\[5ex] = \dfrac{59}{112} \\[5ex] (h.) \\[3ex] n(10^{th}) = 1268 \\[3ex] n(10^{th}\:\:\:AND\:\:\:Math) = 232 \\[3ex] P(Math\:\:from\:\:10^{th}) = \dfrac{n(10^{th}\:\:\:AND\:\:\:Math)}{n(10^{th})} \\[5ex] = \dfrac{232}{1268} \\[5ex] = \dfrac{58}{317} \\[5ex] (i.) \\[3ex] P(Two\:\:Math\:\:from\:\:11^{th}) = P(Math\:\:from\:\:11^{th}) * P(Math\:\:from\:\:11^{th}) ... Multiplication\:\:Rule \\[5ex] P(Two\:\:Math\:\:from\:\:11^{th}) = \dfrac{13}{15} * \dfrac{13}{15} \\[5ex] = \dfrac{13 * 13}{15 * 15} \\[5ex] = \dfrac{169}{225} \\[5ex] (j.) \\[3ex] P(Two\:\:Math\:\:from\:\:10^{th}) = P(Math\:\:from\:\:10^{th}) * P(Math\:\:from\:\:10^{th}) ... Multiplication\:\:Rule \\[5ex] P(Two\:\:Math\:\:from\:\:10^{th}) = \dfrac{58}{317} * \dfrac{58}{317} \\[5ex] = \dfrac{58 * 58}{317 * 317} \\[5ex] = \dfrac{3364}{100489} \\[5ex] (k.) \\[3ex] P(12^{th}\:\:\:OR\:\:\:Math) = P(12^{th}) + P(Math) - P(12^{th}\:\:\:AND\:\:\:Math) ... Addition\:\:Rule \\[3ex] P(12^{th}\:\:\:OR\:\:\:Math) = \dfrac{112}{1890} + \dfrac{733}{1890} - \dfrac{59}{1890} \\[5ex] = \dfrac{112 + 733 - 59}{1890} \\[5ex] = \dfrac{786}{1890} \\[5ex] = \dfrac{131}{315}$
(20.) ACT A batch of 100 defective computer chips consisting of 2 types (I and II) and made by 2 companies (A and B) was selected, and it was determined how many of each type of chip was made by each company.
The results are displayed in the table below.

 Type of chip Number of chips made by Company: $A$ $B$ $I$ $II$ $14$ $36$ $30$ $20$

What is the probability that a randomly selected chip from this batch of $100$ is Type $I$ and manufactured by Company $B?$

$A.\:\: \dfrac{30}{100} \\[5ex] B.\:\: \dfrac{30}{50} \\[5ex] C.\:\: \dfrac{30}{44} \\[5ex] D.\:\: \dfrac{14}{100} \\[5ex] E.\:\: \dfrac{14}{44} \\[5ex]$

Let the event of that a randomly selected chip from this batch of $100$ is Type $I$ and manufactured by Company $B$ = $E$

$n(S) = 100 \\[3ex] n(E) = 30 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{30}{100}...Option\:\:B \\[5ex] = \dfrac{3}{10}$

(21.) About 12% of the population of the U.S state of Alabama smile at strangers.
If two Alabamians are randomly selected:
(a.) What is the probability that both smile at strangers?
(b.) What is the probability that both do not smile at strangers?
(c.) What is the probability that only one smiles at strangers?
(d.) What is the probability that at least one smiles at strangers?

$P(smile) = 12\% = \dfrac{12}{100} = \dfrac{3}{25} \\[5ex] P(smile') = 1 - P(smile) ... Complementary\:\:Rule \\[3ex] P(smile') = 1 - \dfrac{3}{25} = \dfrac{25}{25} - \dfrac{3}{25} = \dfrac{25 - 3}{25} = \dfrac{22}{25} \\[5ex] (a.)\:\:P(both\:\:smile) = \dfrac{3}{25} * \dfrac{3}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] = \dfrac{3 * 3}{25 * 25} = \dfrac{9}{625} \\[5ex] (b.)\:\:P(both\:\:do\:\:not\:\:smile) = \dfrac{22}{25} * \dfrac{22}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] = \dfrac{22 * 22}{25 * 25} = \dfrac{484}{625} \\[5ex] (c.)\:\:P(only\:\:one\:\:smiles) \\[3ex] = P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile\:\:OR\:\:1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) \\[3ex] = P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile) + P(1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) ...Addition\:\:Rule \\[3ex] P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile) = \dfrac{3}{25} * \dfrac{22}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[5ex] P(1st\:\:smiles\:\:AND\:\:2nd\:\:does\:\:not\:\:smile) = \dfrac{3 * 22}{25 * 25} = \dfrac{66}{625} \\[5ex] P(1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) = \dfrac{22}{25} * \dfrac{3}{25} ... Multiplication\:\:Rule - Independent\:\:Events \\[5ex] P(1st\:\:does\:\:not\:\:smile\:\:AND\:\:2nd\:\:smiles) = \dfrac{22 * 3}{25 * 25} = \dfrac{66}{625} \\[5ex] = \dfrac{66}{625} + \dfrac{66}{625} \\[5ex] = \dfrac{66 + 66}{625} \\[5ex] = \dfrac{132}{625} \\[5ex] (d.)\:\:can\:\:be\:\:done\:\:in\:\:two\:\:ways \\[3ex] \underline{First\:\:Method - Less\:\:Work} \\[3ex] P(at\:\:least\:\:one\:\:smiles) = 1 - P(both\:\:do\:\:not\:\:smile) ... Complementary\:\:Rule - Complementary\:\: Events \\[3ex] = 1 - \dfrac{484}{625} \\[5ex] = \dfrac{625}{625} - \dfrac{484}{625} \\[5ex] = \dfrac{625 - 484}{625} \\[5ex] = \dfrac{141}{625} \\[5ex] \underline{Second\:\:Method - More\:\:Work} \\[3ex] P(at\:\:least\:\:one\:\:smiles) = P(only\:\:one\:\:smiles) + P(both\:\:smile) \\[3ex] = \dfrac{132}{625} + \dfrac{9}{625} \\[5ex] = \dfrac{132 + 9}{625} \\[5ex] = \dfrac{141}{625}$
(22.) 11 of the 50 laptops sold in a certain store in the town of Coward, South Carolina on a certain Black Friday were defective.
Determine the probability of randomly selecting a non-defective laptop.

$Let\:\: Defective = D \\[3ex] Non-defective = ND \\[3ex] n(S) = 50 \\[3ex] n(D) = 11 \\[3ex] n(ND) = 50 - 11 = 39 \\[3ex] P(ND) = \dfrac{n(ND)}{n(S)} = \dfrac{39}{50}$
(23.) WASSCE The probabilities that two athletes P and Q will win a gold medal in a competition are 0.75 and 0.60 respectively.
What is the probability that in the competition,
(i) both P and Q will win gold medals,
(ii) neither of them will win a gold medal,
(iii) at least one of them will win a gold medal?

$P(P\:\: wins) = P(P) = 0.75 \\[3ex] P(P\:\: loses) = (P') = 1 - 0.75 = 0.25 \\[3ex] P(Q\:\: wins) = P(Q) = 0.60 \\[3ex] P(Q\:\: loses) = P(Q') = 1 - 0.60 = 0.40 \\[3ex] (i)\:\: P(both\:\: win) \\[3ex] = P(PQ) \\[3ex] = P(P) * P(Q) \\[3ex] = (0.75)(0.6) \\[3ex] = 0.45 \\[5ex] (ii)\:\: P(none\:\:wins) = P(both\:\:lose) \\[3ex] = P(P'Q') \\[3ex] = P(P') * P(Q') \\[3ex] = (0.25)(0.4) \\[3ex] = 0.1 \\[5ex] (iii)\:\: We\:\:can\:\:solve\:\:in\:\:two\:\:ways \\[3ex] \underline{First\:\:Method:\:\: Quantitative\:\:Reasoning} \\[3ex] P(at\:\:least\:\:one\:\:wins) \\[3ex] = P(P\:\:wins\:\:AND\:\:Q\:\:loses) \:\:OR\:\: P(P\:\:loses\:\:AND\:\:Q\:\:wins) \:\:OR\:\: P(P\:\:wins\:\:AND\:\:Q\:\:wins) \\[3ex] = (0.75)(0.4) + (0.25)(0.6) + (0.75)(0.6) \\[3ex] = 0.3 + 0.15 + 0.45 \\[3ex] = 0.9 \\[3ex] \underline{Second\:\:Method:\:\: Complementary\:\: Events} \\[3ex] P(at\:\:least\:\:one\:\:wins) = 1 - P(both\:\: lose) ... Complementary\:\: Rule \\[3ex] = 1 - 0.1 \\[3ex] = 0.9$
(24.) Let the sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Assume all the outcomes are equally likely.
Compute the probability of the event, E = {3, 10}

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\[3ex] n(S) = 10 \\[3ex] E = \{3, 10\} \\[3ex] n(E) = 2 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{2}{10} = \dfrac{1}{5}$
(25.) A drug testing company located in the City of Surprise, Arizona tested several job applicants for marijuana.
146 applicants tested positive, but among them were 26 false positive results.
160 applicants tested negative, but among them were 3 false negative results.
(a.) Represent this information in a table.
(b.) How many job applicants were tested?
(c.) How many applicants did not use marijuana?
(d.) How many applicants used marijuana?
(e.) Determine the probability that a randomly selected applicant did not use marijuana.
(f.) Determine the probability that a randomly selected applicant used marijuana.
(g.) Determine the probability that a randomly selected applicant tested negative OR did not use marijuana.

Recall:
In testing for marijuana and for most applicable tests:

A False Positive means that the applicant did not use it but tested positive for it.

A False Negative means that the applicant used it but tested negative for it.

A True Positive means that the applicant used it and tested positive for it.

A True Negative means that the applicant did not use it and tested negative for it.

$Let\:\: Positive = P \\[3ex] True\:\:Positive = TP \\[3ex] False\:\:Positive = FP \\[3ex] Negative = N \\[3ex] True\:\:Negative = TN \\[3ex] False\:\:Negative = FN \\[3ex] n(P) = 146 \\[3ex] n(FP) = 26 \\[3ex] n(TP) = 146 - 26 = 120 \\[3ex] n(N) = 160 \\[3ex] n(FN) = 3 \\[3ex] n(TN) = 160 - 3 = 157 \\[3ex]$
(a.) Marijuana Use
True False
Positive Result $120$ $26$
Negative Result $157$ $3$

$(b.)\:\:n(S) = n(P) + n(N) \\[3ex] = 146 + 160 \\[3ex] = 306\:\:applicants \\[3ex] (c.)\:\:n(No\:\:Marijuana) = n(TN) + n(FP) \\[3ex] = 157 + 26 \\[3ex] = 183\:\:applicants \\[3ex] (d.)\:\:n(Yes\:\:Marijuana) = n(TP) + n(FN) \\[3ex] = 120 + 3 \\[3ex] = 123\:\:applicants \\[3ex] (e.)\:\:P(No\:\:Marijuana) = \dfrac{n(No\:\:Marijuana)}{n(S)} \\[5ex] = \dfrac{183}{306} \\[5ex] = \dfrac{61}{102} \\[5ex] (f.)\:\:P(Yes\:\:Marijuana) = \dfrac{n(Yes\:\:Marijuana)}{n(S)} \\[5ex] = \dfrac{123}{306} \\[5ex] = \dfrac{41}{102} \\[5ex] (g.)\:\:P(N \cup NM) = P(N) + P(NM) - P(N \cap NM) ... Addition\:\:Rule \\[3ex] P(N) = \dfrac{n(N)}{n(S)} = \dfrac{160}{306} \\[5ex] P(NM) = \dfrac{n(NM)}{n(S)} = \dfrac{183}{306} \\[5ex] P(N \cap NM) = \dfrac{n(N \cap NM)}{n(S)} = \dfrac{n(TN)}{n(S)} = \dfrac{157}{306} \\[5ex] \therefore P(N \cup NM) = \dfrac{160}{306} + \dfrac{183}{306} - \dfrac{157}{306} \\[5ex] = \dfrac{160 + 183 - 157}{306} \\[5ex] = \dfrac{186}{306} \\[5ex] = \dfrac{31}{51}$
(26.) The hours of sleep that citizens get on a typical night in the City of Boring, Oregon is shown in the table below.

Hours of Sleep Number of Citizens (in millions)
$4\:\:or\:\:less$ $11$
$5$ $31$
$6$ $74$
$7$ $89$
$8$ $81$
$9$ $10$
$10\:\:or\:\:more$ $4$

A researcher in sleep deprivation finds that the average human needs at least six hours of sleep a night to function properly.
Determine the probability of a citizen getting at least six hours of sleep a night.

Let the event of finding a citizen who gets at least six hours of sleep a night = $E$

$n(S) = 11 + 31 + 74 + 89 + 81 + 10 + 4 = 300\:\:milion \\[3ex] At\:\:least\:\:6\:\:means\:\: \gt 6 \\[3ex] n(E) = 74 + 89 + 81 + 10 + 4 = 258\:\: million \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{258}{300} = \dfrac{86}{100} = \dfrac{43}{50}$
(27.) ACT If a marble is randomly chosen from a bag that contains exactly 8 red marbles, 6 blue marbles, and 6 white marbles, what is the probability that the marble will NOT be white?

$F.\:\: \dfrac{3}{4} \\[5ex] G.\:\: \dfrac{3}{5} \\[5ex] H.\:\: \dfrac{4}{5} \\[5ex] J.\:\: \dfrac{3}{10} \\[5ex] K.\:\: \dfrac{7}{10} \\[5ex]$

We can solve this in at least two ways
Choose whatever approach you prefer

$Let: \\[3ex] red = R \\[3ex] blue = B \\[3ex] white = W \\[3ex] S = \{8R, 6B, 6W\} \\[3ex] n(S) = 8 + 6 + 6 = 20 \\[5ex] \underline{First\:\: Method: Quantitative\:\: Reasoning} \\[3ex] NOT\:\:White \implies R\:\:AND\:\:B \\[3ex] n(R) = 8 \\[3ex] n(B) = 6 \\[3ex] n(R\:\:AND\:\:B) = 8 + 6 = 14 \\[3ex] P(NOT\:\:White) = P(R\:\:AND\:\:B) = \dfrac{n(R\:\:AND\:\:B)}{n(S)} \\[5ex] P(NOT\:\:White) = \dfrac{14}{20} \\[5ex] P(NOT\:\:White) = \dfrac{7}{10} \\[5ex] \underline{Second\:\: Method: Complementary\:\:Rule} \\[3ex] NOT\:\:white = W' \\[3ex] S = \{8R, 6B, 6W\} \\[3ex] n(S) = 8 + 6 + 6 = 20 \\[3ex] P(White) + P(NOT\:\:White) = 1 ... Complementary\:\: Rule \\[3ex] P(W) + P(W') = 1 \\[3ex] n(W) = 6 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} = \dfrac{6}{20} \\[5ex] P(W') = 1 - P(W) \\[3ex] P(W') = 1 - \dfrac{6}{20} \\[5ex] P(W') = \dfrac{20}{20} - \dfrac{6}{20} \\[5ex] P(W') = \dfrac{20 - 6}{20} \\[5ex] P(W') = \dfrac{14}{20} \\[5ex] P(W') = \dfrac{7}{10}$
(28.) ACT Yulan will use a bag of 30 solid-colored marbles for a game in which each player randomly draws marbles from the bag.
The number of marbles of each color is shown in the table below.

Color Number
Blue
Red
Black
White
Green
$10$
$8$
$6$
$4$
$2$

Yulan will randomly draw $2$ marbles from the bag, one after the other, without replacing the first marble.
What is the probability that Yulan will draw a black marble first and a green marble second?

$A.\;\; \dfrac{1}{75} \\[5ex] B.\;\; \dfrac{2}{145} \\[5ex] C.\;\; \dfrac{4}{15} \\[5ex] D.\;\; \dfrac{39}{145} \\[5ex] E.\;\; \dfrac{2}{5} \\[5ex]$

$n(S) = 10 + 8 + 6 + 4 + 2 = 30 \\[3ex] Let\:\: Black = B \\[3ex] n(B) = 6 \\[3ex] Green = G \\[3ex] n(G) = 2 \\[3ex] \underline{Without \:\: Replacement \implies Dependent\:\: Events} \\[3ex] P(B \:\:AND\:\: B) = P(BG) \\[3ex] P(BG) = \dfrac{6}{30} * \dfrac{2}{29}...Multiplication\:\: Rule \\[5ex] P(BG) = \dfrac{1}{5} * \dfrac{2}{29} \\[5ex] P(BG) = \dfrac{2}{145}$
(29.) ACT The stem-and-leaf plot below shows the number of rebounds a basketball player with the Connecticut Suns grabbed in each of 17 games.

$$\begin{array}{c|c c} Stem & Leaf \\ \hline 0 & 8 & 9 \\ 1 & 1 & 1 & 3 & 3 & 4 & 4 & 5 & 5 & 5 & 6 & 9 \\ 2 & 1 & 1 & 2 & 4 \end{array}$$ (Note: For example, $13$ rebounds would have a stem value of $1$ and a leaf value of $3$)

$n(S) = 17 \\[3ex] n(15\:\: rebounds) = 3 \\[3ex] P(15\:\: rebounds) = \dfrac{n(15\:\: rebounds)}{n(S)} = \dfrac{3}{17}$
(30.) A bag of 100 marbles contains 35 red marbles, 30 yellow marbles, and 35 purple marbles.
What is the probability that a randomly selected marble is purple?

$Let\:\: red = R \\[3ex] yellow = Y \\[3ex] purple = P \\[3ex] n(S) = 100 \\[3ex] n(P) = 35 \\[3ex] P(P) = \dfrac{n(P)}{n(S)} \\[5ex] P(P) = \dfrac{35}{100} \\[5ex] P(P) = \dfrac{7}{20}$
(31.) ACT A bag contains several marbles.
On 3 successive draws with replacement, a red marble is drawn from the bag each time.
Which of the following statements must be true about the marbles in the bag?

F. At least 1 marble is red.
G. Exactly 1 marble is red.
H. Exactly 3 marbles are red.
J. All the marbles are red.
K. The bag contains more red marbles than marbles of other colors.

Let us analyze each of these options.
F. At least $1$ marble is red.
At least $1$ means "$1$ or more", $\ge 1$
This is a correct option because it is possible that there are one or more red marbles in the bag.
But, let us look at the other options before making a conclusion.

G. Exactly $1$ marble is red.
We are not really sure of this option.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than only $1$ is red.
At least $1$ means $1$ or more
That also includes $1$

H. Exactly $3$ marbles are red.
We are not sure of this option either.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than only $3$ are red.
At least $1$ means $1$ or more
That also includes $3$

J. All the marbles are red.
We are not sure of this option as well.
What if only $2$ marbles are red? It is still possible to pick a red marble three times in a row with replacement in a bag of several marbles if only $2$ marbles are red.
It is much better to go with the first option (at least $1$ are red) rather than $all$ are red.
At least $1$ means $1$ or more
That also includes $all\:\: the\:\: marbles$

K. The bag contains more red marbles that marbles of other colors.
We are not sure of this option.
What if the bag contained only red marbles?
What if the bag contained more green marbles than red marbles and one just picked only a red marble each time because of the "with replacement" condition?
So, it is not okay to go with this option given the nature of the question.

The correct option is F. At least $1$ marble is red.
(32.) Which of the following numbers could be the probability of an event?

$1, 1.37, 0.26, -0.52, 0.05, 0, \dfrac{25}{12}, 120\%$

$1:\;\; YES \\[3ex] 1.37:\;\; NO \\[3ex] 0.26:\;\; YES \\[3ex] -0.52:\;\; NO \\[3ex] 0.05:\;\; YES \\[3ex] 0:\;\; YES \\[3ex] \dfrac{25}{12} = 2.08333:\;\; NO \\[5ex] 120\% = \dfrac{120}{100} = 1.2:\;\; NO \\[5ex]$ The probabilities of events could be: $0, 0.05, 0.26, 1$
The probability of an impossible event is $0$
The probability of an event that must occur (a surely certainty event) is $1$
The probability of any event is from $0$ through $1$
(33.) Does the table represent a probability model?
How would you describe the event of selecting a brown color?

Color Probability
red $0.25$
green $0.2$
blue $0.15$
brown $0$
yellow $0.35$
orange $0.1$

$\Sigma Probability = 1.05 \\[3ex] \Sigma Probability \ne 1 \\[3ex]$ Therefore, it does not represent a probability model.
The event of selecting a brown color is an impossible event.
(34.) A multiple-choice test has five possible answers.
However, only one answer is correct for each question.
Determine the probability of:

$n(S) = 5 \\[3ex] n(correct\:\: answer) = 1 \\[3ex] n(incorrect\:\: answers) = 5 - 1 = 4 \\[3ex] P(correct\:\: answer) = \dfrac{n(correct\:\: answer)}{n(S)} = \dfrac{1}{5} \\[5ex] P(incorrect\:\: answers) = \dfrac{n(incorrect\:\: answers)}{n(S)} = \dfrac{4}{5}$
(35.) A loaded die is a die in which a certain outcome is more likely.
A fair die is a die in which all outcomes are equally likely.
A die is rolled 400 times.
The outcome of the experiment is listed in the table below.

Value of Die Frequency, $F$
$1$ $71$
$2$ $62$
$3$ $66$
$4$ $63$
$5$ $70$
$6$ $68$
$\Sigma F = 400$

The die is not loaded (it is a fair die) because each value has an approximately equal chance of occurrence.
The frequency of the values are very close to each other.
(36.) ACT If a bag contains 5 blue marbles, 4 red marbles, and 3 green marbles, what is the probability that a marble randomly picked from the bag will be red?

$F.\:\: \dfrac{1}{12} \\[5ex] G.\:\: \dfrac{1}{4} \\[5ex] H.\:\: \dfrac{1}{3} \\[5ex] J.\:\: \dfrac{5}{12} \\[5ex] K.\:\: \dfrac{2}{3} \\[5ex]$

$Let \\[3ex] blue\:\:marble = B \\[3ex] red\:\:marble = R \\[3ex] green\:\:marble = G \\[3ex] S = \{5B, 4R, 3G\} \\[3ex] n(S) = 5 + 4 + 3 = 12 \\[3ex] n(R) = 4 \\[3ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{4}{12} \\[5ex] P(R) = \dfrac{1}{3}$
(37.) A loaded die is a die in which a certain outcome is more likely.
A fair die is a die in which all outcomes are equally likely.
A die is rolled 400 times.
The outcome of the experiment is listed in the table below.

Value of Die Frequency, $F$
$1$ $43$
$2$ $100$
$3$ $44$
$4$ $114$
$5$ $50$
$6$ $49$
$\Sigma F = 400$

The die is loaded (it is a loaded die) because each value does not have an approximately equal chance of occurrence.
The frequency of the values of $2$ and $4$ are very high.
(38.) JAMB The letters of the word MATRICULATION are cut and put into a box.
One of the letters is drawn at random from the box.
Find the probability of drawing a vowel.

$A.\:\: \dfrac{2}{13} \\[5ex] B.\:\: \dfrac{5}{13} \\[5ex] C.\:\: \dfrac{6}{13} \\[5ex] D.\:\: \dfrac{8}{13} \\[5ex] E.\:\: \dfrac{4}{13} \\[5ex]$

$MATRICULATION \\[3ex] n(S) = 13 \\[3ex] Vowels = \{A,I,U,A,I,O\} \\[3ex] n(Vowels) = 6 \\[3ex] P(Vowels) = \dfrac{n(Vowels)}{n(S)} \\[5ex] P(Vowels) = \dfrac{6}{13}$
(39.) WASSCE The probabilities that Mensah will pass a Mathematics and an Economics tests are $\dfrac{3}{4}$ and $\dfrac{5}{8}$ respectively.
If the probability that he passes at least one of the subjects is $\dfrac{7}{12}$, what is the probability that he passes both subjects?

Based on the fact that they gave the conditional statement: If the probability that he passes at least one of the subjects is $\dfrac{7}{12}$
We have to use it.
At least one means one or more.
In this case, "at least one" means "one or both"
The probability that he passes at least one of the subjects means that:
Either he passes Mathematics AND does not pass Economics OR
He passes Economics AND does not pass Mathematics OR
He passes both subjects...he passes both Mathematics and Economics

$Let: \\[3ex] pass\:\:Mathematics = M \\[3ex] P(M) = \dfrac{3}{4} \\[5ex] does\:\:not\:\:pass\:\:Mathematics = M' \\[3ex] P(M') = 1 - \dfrac{3}{4}...Complementary\:\:Rule \\[3ex] = \dfrac{4}{4} - \dfrac{3}{4} \\[5ex] = \dfrac{4 - 3}{4} \\[5ex] = \dfrac{1}{4} \\[5ex] pass\:\:Economics = E \\[3ex] P(E) = \dfrac{5}{8} \\[5ex] does\:\:not\:\:pass\:\:Economics = E' \\[3ex] P(E') = 1 - \dfrac{5}{8}...Complementary\:\:Rule \\[5ex] = \dfrac{8}{8} - \dfrac{5}{8} \\[5ex] = \dfrac{8 - 5}{8} \\[5ex] = \dfrac{3}{8} \\[5ex] pass\:\:both = ME...what\:\:we\:\:need\:\:to\:\:find \\[3ex] P(ME') = P(M) * P(E')...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] = \dfrac{3}{4} * \dfrac{3}{8} \\[5ex] = \dfrac{3 * 3}{4 * 8} \\[5ex] = \dfrac{9}{32} \\[5ex] P(EM') = P(E) * P(M')...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] = \dfrac{5}{8} * \dfrac{1}{4} \\[5ex] = \dfrac{5 * 1}{8 * 4} \\[5ex] = \dfrac{5}{32} \\[5ex] P(at\:\:least\:\:one\:\:of\:\:the\:\:subjects) = \dfrac{7}{12} \\[5ex] P(at\:\:least\:\:one\:\:of\:\:the\:\:subjects) = P(ME') + P(EM') + P(ME) \\[3ex] \implies \dfrac{7}{12} = \dfrac{9}{32} + \dfrac{5}{32} + P(ME) \\[5ex] \dfrac{9}{32} + \dfrac{5}{32} + P(ME) = \dfrac{7}{12} \\[5ex] \dfrac{9 + 5}{32} + P(ME) = \dfrac{7}{12} \\[5ex] \dfrac{14}{32} + P(ME) = \dfrac{7}{12} \\[5ex] \dfrac{7}{16} + P(ME) = \dfrac{7}{12} \\[5ex] P(ME) = \dfrac{7}{12} - \dfrac{7}{16} \\[5ex] = \dfrac{28}{48} - \dfrac{21}{48} \\[5ex] = \dfrac{28 - 21}{48} \\[5ex] = \dfrac{7}{48} \\[5ex]$ The probability that he passes both subjects is $\dfrac{7}{48}$
(40.) With one method of a procedure called acceptance sampling, a sample of items is randomly selected without replacement and the entire batch is rejected if there is at least one defective unit.
Prosperity Electronics Company has just manufactured 5000 write-rewrite CDs, and $180$ are defective.
If 4 of these CDs are randomly selected and tested, what is the probability that the entire batch will be rejected?

We can solve this question in at least two ways.
It is highly recommended that you use the Second Method...using the Complementary Rule

At least one means one or more
This means that out of the 4 randomly selected CDS,
Reject the entire batch if at least one of the CDs is defective.
This means that Reject the entire batch if:
the 1st is defective AND the 2nd is not AND the 3rd is not AND the 4th is not
OR
the 1st is defective AND the 2nd is defective AND the 3rd is not AND the 4th is not
OR
the 1st is defective AND the 2nd is defective AND the 3rd is defective AND the 4th is not
OR
All four are defective
OR
the 1st is not AND the 2nd is defective AND the 3rd is not AND the 4th is not
OR
the 1st is not AND the 2nd is not AND the 3rd is defective AND the 4th is not
OR...and the possibilities keep going...
It is a Herculean task to use this method.

So, we shall use the Complementary Rule
(At least one is defective) is 1 minus (none are defective)

$Let \\[3ex] Defective\:\:CD = D \\[3ex] Non-defective\:\:CD = D' \\[3ex] n(S) = 5000 \\[3ex] n(D) = 180 \\[3ex] n(D') = 5000 - 180 = 4820 \\[3ex] \underline{2nd\:\:Method:\:\:Complementary\:\:Rule} \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) \\[3ex] = 1 - P(none\:\:are\:\:defective) \\[3ex] = 1 - P(D'D'D'D') \\[3ex] P(D'D'D'D') = \dfrac{4820}{5000} * \dfrac{4820}{5000} * \dfrac{4820} * \dfrac{4820}{5000} \\[5ex] P(D'D'D'D') = \left(\dfrac{4820}{5000}\right)^4 \\[5ex] P(D'D'D'D') = 0.964^4 \\[3ex] P(D'D'D'D') = 0.8635910556 \\[3ex] \implies P(at\:\:least\:\:one\:\:is\:\:defective) = 1 - 0.8635910556 \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) = 0.1364089444 \\[3ex] converting\:\:to\:\:percent \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) = 0.1364089444 * 100 \\[3ex] P(at\:\:least\:\:one\:\:is\:\:defective) = 13.64089444\%$

(41.) A group of airline passengers responded to a question about their favorite seat on a plane.
501 passengers chose the window seat, 10 chose the middle seat, and 297 chose the aisle seat.
(a.) Determine the probability that a passenger prefers the middle seat.

(b.) Is it unlikely for a passenger to prefer the middle seat? Why?

$(a.) \\[3ex] Let \\[3ex] Window\:\:seat = W \\[3ex] Middle\:\:seat = M \\[3ex] Aisle\:\:seat = A \\[3ex] n(W) = 501 \\[3ex] n(M) = 10 \\[3ex] n(A) = 297 \\[3ex] n(S) = 501 + 10 + 297 = 808 \\[3ex] P(M) = \dfrac{n(M)}{n(S)} = \dfrac{10}{808} = 0.0123762376 \\[5ex] P(M) \approx 0.012 \\[3ex] (b.) \\[3ex]$ Rare Event Rule
Yes, it is unlikely for a passenger to prefer the middle seat because the probability that a passenger prefers the middle seat, $0.012$ is less than $0.05$
This may be due to the fact the middle seat lacks easy access to the aisle and the outside view among others.
(42.) JAMB Bola chooses at random a number between 1 and 300.
What is the probability that the number is divisible by 4?

$A.\:\: \dfrac{1}{3} \\[5ex] B.\:\: \dfrac{1}{4} \\[5ex] C.\:\: \dfrac{1}{5} \\[5ex] D.\:\: \dfrac{4}{300} \\[5ex] E.\:\: \dfrac{1}{300} \\[5ex]$

Divisibility by Numbers
A number is divisible by $4$ if it is an even number AND if the last two digits are divisible by $4$
Between $1$ and $300$, there are $75$ numbers that are divisible by $4$

Student: Samdom For Peace, how did you know?
Why not list them and count them?
Teacher: Good question.
This is a JAMB question.
You should solve this question accurately on time without a calculator
Between $1$ and $12$, list the numbers that are divisible by $4$
Student: The numbers are: $4, 8, 12$
Teacher: How many are there?
Student: Three numbers, Sir
Teacher: Correct!
Between $1$ and $20$, list the numbers that are divisible by $4$
Student: The numbers are: $4, 8, 12, 16, 20$
Teacher: How many are there?
Student: Five numbers, Sir
Teacher: Correct!
Student: Okay, I understand

$12 \div 4 = 3 \\[3ex] 20 \div 4 = 5 \\[3ex] 300 \div 4 = 75 \\[3ex]$
Let the set of numbers that are divisible by $4 = E$

$n(E) = 75 \\[3ex] n(S) = 300 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{75}{300} \\[5ex] P(E) = \dfrac{1}{4}$
(43.) JAMB What is the probability that a number chosen at random from the integers between 1 and 10 inclusive is either a prime or a multiple of 3?

$A.\:\: \dfrac{7}{10} \\[5ex] B.\:\: \dfrac{3}{5} \\[5ex] C.\:\: \dfrac{4}{5} \\[5ex] D.\:\: \dfrac{1}{2} \\[5ex] E.\:\: \dfrac{3}{10} \\[5ex]$

$OR \implies \cup \\[3ex] AND \implies \cap \\[3ex] S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\} \\[3ex] n(S) = 10 \\[3ex] Let\:\:A\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:prime \\[3ex] A = \{2, 3, 5, 7\} \\[3ex] n(A) = 4 \\[3ex] P(A) = \dfrac{n(A)}{n(S)} = \dfrac{4}{10} \\[5ex] Let\:\:B\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:multiple\:\:of\:\:3 \\[3ex] B = \{3, 6, 9\} \\[3ex] n(B) = 3 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} = \dfrac{3}{10} \\[5ex] A \cap B = \{3\} \\[3ex] n(A \cap B) = 1 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(S)} = \dfrac{1}{10} \\[5ex] P(A \cup B) = P(A) + P(B) - P(A \cap B)...Addition\:\:Rule \\[3ex] P(A \cup B) = \dfrac{4}{10} + \dfrac{3}{10} - \dfrac{1}{10} \\[5ex] P(A \cup B) = \dfrac{4 + 3 - 1}{10} \\[5ex] P(A \cup B) = \dfrac{6}{10} \\[5ex] P(A \cup B) = \dfrac{3}{5} \\[5ex]$ Between the numbers $1$ and $10$ inclusive, there is a $\dfrac{3}{5}$ probability of picking a prime number or a multiple of $3$
(44.) ACT Martin has an empty bag and puts in 3 red marbles.
He now wants to put in enough green marbles so the probability of drawing a red marble at random from the bag is $\dfrac{1}{4}$
How many green marbles should he put in?

$A.\:\: 1 \\[3ex] B.\:\: 3 \\[3ex] C.\:\: 5 \\[3ex] D.\:\: 9 \\[3ex] E.\:\: 12 \\[3ex]$

We can solve this question in two ways.
Use any method you prefer

Let the number of green marbles to put in the bag = $x$

$Let\:\: red = R \\[3ex] n(R) = 3 \\[3ex] green = G \\[3ex] \underline{First\:\:Method} \\[3ex] n(G) = x \\[3ex] n(S) = 3 + x \\[3ex] P(R) = \dfrac{n(R)}{n(S)} = \dfrac{3}{3 + x} \\[5ex] Also\:\: P(R) = \dfrac{1}{4}...from\:\:the\:\:question \\[5ex] \rightarrow \dfrac{3}{3 + x} = \dfrac{1}{4} \\[5ex] Cross\:\:Multiply \\[3ex] 1(3 + x) = 3(4) \\[3ex] 3 + x = 12 \\[3ex] x = 12 - 3 \\[3ex] x = 9 \\[3ex] \underline{Second\:\:Method} \\[3ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{1}{4} \\[5ex] \rightarrow \dfrac{3}{n(S)} = \dfrac{1}{4} \\[5ex] Cross\:\:Multiply \\[3ex] n(S) * 1 = 3(4) \\[3ex] n(S) = 12 \\[3ex] n(G) = 12 - 3 \\[3ex] n(G) = 9 \\[3ex]$ Martin should put $9$ green marbles in the bag.

$\underline{Check} \\[3ex] 3 + x = 3 + 9 = 12 \\[3ex] P(R) = \dfrac{3}{3 + x} = \dfrac{3}{12} = \dfrac{1}{4}$
(45.) According to the Milk Marketing Board of the State of Pennsylvania, United States of America, milk classification falls into four categories: Class I, Class, II, Class III, and Class IV.

Suppose 88% of milk falls in the Class I category.

(a.) Two containers of milk are selected at random. Determine the probability that both contain Class I milk.

(b.) Four containers of milk are selected at random. What is the probability that four of them DO NOT Class I milk.
Write an integer or decimal.

$Let:\:\:Class\:\:I\:\:milk = M \\[3ex] \implies NOT\:\:Class\:\:I\:\:milk = M' \\[3ex] P(M) = 88\% = \dfrac{88}{100} = 0.88 \\[5ex] P(M) + P(M') = 1 ...Complementary\:\:Events \\[3ex] P(M') = 1 - P(M) \\[3ex] P(M') = 1 - 0.88 \\[3ex] P(M') = 0.12 \\[3ex] (a.) \\[3ex] P(MM) = (0.88)(0.88) ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(MM) = 0.7744 \\[3ex] (b.) \\[3ex] P(M'M'M'M') = (0.12)(0.12)(0.12)(0.12) = 0.12^4 ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(M'M'M'M') = 0.00020736 \\[3ex]$ (a.) The probability that two containers of milk selected at random contain Class $I$ milk is $0.7744$
(b.) The probability that four containers of milk selected at random DO NOT contain Class $I$ milk is $0.00020736$
(46.) JAMB Two numbers are chosen at random from three numbers 1, 3, 6.
Find the probability that the sum of the two is not odd.

$A.\:\: \dfrac{2}{3} \\[5ex] B.\:\: \dfrac{1}{2} \\[5ex] C.\:\: \dfrac{1}{3} \\[5ex] D.\:\: \dfrac{1}{6} \\[5ex]$

$Sample\:\:Space \\[3ex] 1 + 3 = 4 \\[3ex] 1 + 6 = 7 \\[3ex] 3 + 6 = 9 \\[3ex] S = \{4, 7, 9\} \\[3ex] n(S) = 3 \\[3ex] NOT\:\:Odd = \{4\} \\[3ex] n(NOT\:\:Odd) = 1 \\[3ex] P(NOT\:\:Odd) = \dfrac{n(NOT\:\:Odd)}{n(S)} \\[5ex] P(NOT\:\:Odd) = \dfrac{1}{3}$
(47.) ACT A science class has 8 juniors and 4 seniors.
The teacher will randomly select 2 students, one at a time, to represent the class in a committee at the school.
Given that the first student selected is a junior, what is then probability that the second student selected will be a senior?

$A.\:\: \dfrac{1}{11} \\[5ex] B.\:\: \dfrac{1}{4} \\[5ex] C.\:\: \dfrac{3}{11} \\[5ex] D.\:\: \dfrac{1}{3} \\[5ex] E.\:\: \dfrac{4}{11} \\[5ex]$

This is a case of Conditional Probability
Let us reword the question:
What is the probability of selecting a senior given that a junior was selected

$n(juniors) = 8 \\[3ex] n(seniors) = 4 \\[3ex] n(S) = 8 + 4 = 12 \\[3ex] P(junior) = \dfrac{n(juniors)}{n(S)} = \dfrac{8}{12} = \dfrac{2}{3} \\[5ex] P(senior) = \dfrac{n(seniors)}{n(S)} = \dfrac{4}{12} = \dfrac{1}{3} \\[5ex] P(senior | junior) = \dfrac{P(senior\:\:AND\:\:junior)}{P(junior)} \\[5ex] P(senior\:\:AND\:\:junior) = P(senior) * P(junior)...Multiplication\:\:Rule\:\:for\:\:Dependent\:\:Events \\[3ex] = \dfrac{4}{12} * \dfrac{8}{11}...Without\:\:Replacement \\[5ex] = \dfrac{1}{3} * \dfrac{8}{11} \\[5ex] = \dfrac{8}{33} \\[5ex] \implies P(senior | junior) = P(senior\:\:AND\:\:junior) \div P(junior) \\[3ex] = \dfrac{8}{33} \div \dfrac{2}{3} \\[5ex] = \dfrac{8}{33} * \dfrac{3}{2} \\[5ex] = \dfrac{4}{11}$
(48.) ACT A wallet containing 2 five-dollar bills, 9 ten-dollar bills, and 5 twenty-dollar bills is found and returned to its owner.
The wallet's owner will reward the finder with 1 bill drawn randomly from the wallet.
What is the probability that the bill drawn will be a twenty-dollar bill?

$F.\:\: \dfrac{1}{16} \\[5ex] G.\:\: \dfrac{1}{10} \\[5ex] H.\:\: \dfrac{1}{5} \\[5ex] J.\:\: \dfrac{5}{16} \\[5ex] K.\:\: \dfrac{5}{11} \\[5ex]$

$Let: \\[3ex] five-dollar\:\:bill = F \\[3ex] n(F) = 2 \\[3ex] ten-dollar\:\:bill = T \\[3ex] n(T) = 9 \\[3ex] twenty-dollar\:\:bill = W \\[3ex] n(W) = 5 \\[3ex] S = \{2F, 9T, 5W\} \\[3ex] n(S) = 2 + 9 + 5 = 16 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] P(W) = \dfrac{5}{16}$
(49.) Indicate whether this case is true positive, true negative, false positive, or false negative.

Case: Court Verdict: Suspect A murdered her daughter.
The murder was ruled as death by undermined means.
The prosecution team could not provide convincing evidence that she was guilty.
The defense team convinced the jury that she was innocent.
The jury acquitted her.

This is a case of false negative
Suspect $A$ actually murdered her daughter.
However, because the prosecution team could not provide substantial evidence that she murdered her daughter, the jury acquitted her.

NOTE: This is similar to Type $II$ error in Statistics (Errors in Hypothesis Tests): Acquitting a guilty person
(50.) Indicate whether this case is true positive, true negative, false positive, or false negative.

Case: Secondhand Marijuana Smoke: Augustine hangs out frequently with a friend who smokes pot.
THC (Tetrahydrocannbinol) is a chemical found in pot (cannabis plants).
A urine test done immediately after he visited his friend showed traces of THC.
He vehemently denied smoking pot and requested for a second test.

This is a case of false positive
This means that he did not smoke cannabis but tested positive for it.
Augustine did not actually smoke cannabis but tested positive for it because he probably inhaled it from his friend.
(51.) Indicate whether this case is true positive, true negative, false positive, or false negative.

Case: Court Verdict: A record-breaking number of people were exonerated in 2015 — freed after serving time in American prisons for crimes they did not commit. - NBC News - February 3, 2016
One such case was that of William Vasquez who was cleared in December of an arson that killed a mother and her five children in 1981.

This is a case of false positive
This means that he did not commit the crime but was jailed for it.

NOTE: This is similar to Type $I$ error in Statistics (Errors in Hypothesis Tests): Convicting an innocent person
(52.) Indicate whether this case is true positive, true negative, false positive, or false negative.

Case: Early Menstrual Period: Menstruation usually occurs on average of $14$ days after ovulation.
However, Felicity noticed an early menstrual period and decided to take a home pregnancy test.
The test was negative.
But, she was convinced that she was pregant.
So, she waited a few days and tested again in the morning.

This is a case of false negative
Felicty was pregnant.
However, she probably did not have a high enough blood and urine pregnancy hormone (HCG - Human Chorionic Gonadotropin) level to get a positive pregnancy test.
(53.) ACT Five balls, numbered 1, 2, 3, 4, and 5, are placed in a bin.
Two balls are drawn at random without replacement.
What is the probability that the sum of the numbers on the balls drawn is 7?

$F.\:\: \dfrac{1}{5} \\[5ex] G.\:\: \dfrac{2}{5} \\[5ex] H.\:\: \dfrac{4}{5} \\[5ex] J.\:\: \dfrac{5}{9} \\[5ex] K.\:\: \dfrac{4}{25} \\[5ex]$

It is best to represent this as a table.
But, please keep in mind that this is without replacement
So, we will need to delete the sum of the repeated numbers such as:

$(1, 1 = 1 + 1 = 2), \\[3ex] (2, 2 = 2 + 2 = 4), \\[3ex] (3, 3 = 3 + 3 = 6), \\[3ex] (4, 4 = 4 + 4 = 8), \\[3ex] (5, 5 = 5 + 5 = 10) \\[3ex]$ The highlighted numbers in the table are omitted ...because the two balls are drawn without replacement.
The numbers in red color (the $7's$) are the numbers in the event space.

$(+)$ $1$ $2$ $3$ $4$ $5$
$1$ $2$ $3$ $4$ $5$ $6$
$2$ $3$ $4$ $5$ $6$ $\color{red}{7}$
$3$ $4$ $5$ $6$ $\color{red}{7}$ $8$
$4$ $5$ $6$ $\color{red}{7}$ $8$ $9$
$5$ $6$ $\color{red}{7}$ $8$ $9$ $10$

$n(S) = 25 - 5 = 20 \\[3ex] n(7) = 4 \\[3ex] P(7) = \dfrac{n(7)}{n(S)} \\[5ex] P(7) = \dfrac{4}{20} \\[5ex] P(7) = \dfrac{1}{5}$
(54.) JAMB A container has 30 gold medals, 22 silver medals and 18 bronze medals.
If one medal is selected at random from the container, what is the probability that it is not a gold medal?

$A.\:\: \dfrac{4}{7} \\[5ex] B.\:\: \dfrac{3}{7} \\[5ex] C.\:\: \dfrac{11}{35} \\[5ex] D.\:\: \dfrac{9}{35} \\[5ex]$

$Let: \\[3ex] gold = G \\[3ex] silver = L \\[3ex] bronze = B \\[3ex] \underline{First\;\;Approach} \\[3ex] n(G) = 30 \\[3ex] n(L) = 22 \\[3ex] n(B) = 18 \\[3ex] n(S) = 30 + 22 + 18 = 70 \\[3ex] n(G') = 70 - 30 = 40 \\[5ex] \underline{Second\;\;Approach} \\[3ex] n(G') = 22 + 18 = 40 \\[3ex] P(G') = \dfrac{n(G')}{n(S)} \\[5ex] P(G') = \dfrac{40}{70} \\[5ex] P(G') = \dfrac{4}{7}$
(55.) ACT Lena pick 1 card at random form a pack of 25 baseball cards.
Each card features the fielding position for 1 of 25 different baseball players.
Each player in the pack has only 1 fielding position.
The table below lists the frequency of fielding positions in the pack.
What is the probability that the card Lena picks will feature an outfielder or a pitcher?

 Fielding position Frequency Catcher Infielder Pitcher Outfielder $4$ $6$ $8$ $7$

$F.\:\: 9\% \\[3ex] G.\:\: 28\% \\[3ex] H.\:\: 32\% \\[3ex] J.\:\: 56\% \\[3ex] K.\:\: 60\% \\[3ex]$

$n(S) = \Sigma F = 4 + 6 + 8 + 7 = 25 \\[3ex] n(outfielder) = 7 \\[3ex] P(outfielder) = \dfrac{n(outfielder)}{n(S)} \\[5ex] P(outfielder) = \dfrac{7}{25} \\[5ex] n(pitcher) = 8 \\[3ex] P(pitcher) = \dfrac{n(pitcher)}{n(S)} \\[5ex] P(pitcher) = \dfrac{8}{25} \\[5ex] P(outfielder\:\:OR\:\:\:pitcher) = P(outfielder) + P(pitcher) - P(outfielder-pitcher)...Multiplication\:\:Rule \\[3ex] P(outfielder-pitcher) = 0 ...Mutually\:\:Exclusive\:\:Event \\[3ex] P(outfielder\:\:OR\:\:\:pitcher) = \dfrac{7}{25} + \dfrac{8}{25} \\[5ex] = \dfrac{7 + 8}{25} \\[5ex] = \dfrac{15}{25} \\[5ex] to\:\:percent = \dfrac{15}{25} * 100 \\[5ex] = 15(4) \\[3ex] = 60\%$
(56.) GCSE The four candidates in an election were A, B, C and D
The pie chart shows the proportion of votes for each candidate.

Work out the probability that a person who voted, chosen at random, voted for C

$n(S) = 360^\circ...\angle s \;\;in\;\;a\;\;circle \\[3ex] n(S) = 90 + x + 2x + (2x + 10) \\[3ex] n(S) = 90 + 3x + 2x + 10 \\[3ex] n(S) = 5x + 100 \\[3ex] n(S) = n(S) \\[3ex] \implies 5x + 100 = 360 \\[3ex] 5x = 360 - 100 \\[3ex] 5x = 260 \\[3ex] x = \dfrac{260}{5} \\[5ex] x = 52^\circ \\[3ex] n(C) = 2x + 10 \\[3ex] n(C) = 2(52) + 10 \\[3ex] n(C) = 104 + 10 \\[3ex] n(C) = 114^\circ \\[3ex] P(C) = \dfrac{n(C)}{n(S)} \\[5ex] P(C) = \dfrac{114}{360} \\[5ex] P(C) = \dfrac{19}{60}$
(57.) ACT A physical education teacher recorded the distances, in inches, that her students jumped during a long jump lesson.
The distances of 1 jump by each of the students are represented in the stem-and-leaf below.

$$\begin{array}{c|c c} Stem & Leaf \\ \hline 3 & 7 & 8 \\ 4 & 3 & 5 & 6 & 7 \\ 5 & 2 & 4 & 5 & 8 & 9 \\ 6 & 0 & 1 & 2 & 3 & 6 \\ 7 & 0 & 1 & 2 \end{array}$$ Key: $5 | 2 = 52$ inches

What is the probability that a student chosen at random from the class will have jumped at least $60$ inches?

$F.\:\: \dfrac{5}{24} \\[5ex] G.\:\: \dfrac{8}{24} \\[5ex] H.\:\: \dfrac{5}{19} \\[5ex] J.\:\: \dfrac{7}{19} \\[5ex] K.\:\: \dfrac{8}{19} \\[5ex]$

At least $60$ inches means $60$ inches or more

Let $E$ be the event that a randomly chosen student jumped at least $60$ inches

$E = \{60, 61, 62, 63, 66, 70, 71, 72\} \\[3ex] n(E) = 8 \\[3ex] OR \\[3ex] n(E) = 5 + 3 = 8 \\[3ex] n(S) = 2 + 4 + 5 + 5 + 3 = 19 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{8}{19}$
(58.) ACT The Harrisburg Recreation Center recently changed its hours to open 1 hour later and close 3 hours later than it had previously.
Residents of Harrisburg age 16 or older were given a survey, and 560 residents replied.
The survey asked each resident his or her student status (high school, college, or nonstudent) and what he or she thought about the change in hours (approve, disapprove, or no opinion).
The results are summarized in the table below.

 Student status Approve Disapprove No opinion High school College Nonstudent $30$ $14$ $85$ $4$ $10$ $353$ $11$ $6$ $47$ Total $129$ $367$ $64$

Suppose a person will be chosen at random from these $560$ residents.
Which of the following values is closest to the probability that the person chosen will NOT be a high school student and will NOT have replied with no opinion?

$A.\:\: 0.06 \\[3ex] B.\:\: 0.09 \\[3ex] C.\:\: 0.44 \\[3ex] D.\:\: 0.83 \\[3ex] E.\:\: 0.98 \\[3ex]$

The person chosen will NOT be a high school student and will NOT have replied with no opinion means that the person is a college student who approved or disapproved OR the person is a nonstudent who approved or disapproved. We shall add all of them.
Let $E$ be the event of selecting such a person.
This includes:
All college students who approved = $14$
All college students who disapproved = $10$
All nonstudents who approved = $85$
All nonstudents who disapproved = $353$

$n(E) = 14 + 10 + 85 + 353 = 462 \\[3ex] n(S) = 560 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{462}{560} \\[5ex] P(E) = 0.825 \\[3ex] P(E) \approx 0.83$
(59.) CSEC The marks obtained by 10 students in a test, scored out of 60, are shown below.

 $29$ $38$ $26$ $42$ $38$ $45$ $35$ $37$ $38$ $31$

For the data above, determine the probability that a student chosen at random scores less than half the total marks in the test.

$Total\:\:marks = 60 \\[3ex] Half\:\:of\:\:it = \dfrac{60}{2} = 30 \\[5ex] n(students) = 10 \\[3ex] n(students\:\:who\:\:scored\:\:less\:\:than\:\:30) = 2...the\:\:students\:\:who\:\:scored\:\:29\:\:and\:\:26 \\[3ex] P(students\:\:who\:\:scored\:\:less\:\:than\:\:30) = \dfrac{n(students\:\:who\:\:scored\:\:less\:\:than\:\:30)}{n(students)} \\[5ex] = \dfrac{2}{10} \\[5ex] = \dfrac{1}{5}$
(60.) NSC A bag contains 7 yellow balls, 3 red balls and 2 blue balls.
A ball is chosen at random from the bag and not replaced.
A second ball is then chosen.
Determine the probability that of the two balls chosen, one is red and the other is blue.

Of the two balls chosen, one is red and the other is blue

This means it could be in any order
So, it means that the:
first ball is red AND the second ball is blue OR the
first ball is blue AND the second ball is red

$Let: \\[3ex] yellow = Y \\[3ex] red = R \\[3ex] blue = B \\[3ex] n(R) = 3 \\[3ex] n(B) = 2 \\[3ex] n(S) = 7 + 3 + 2 = 12 \\[3ex] \underline{Without\:\:Replacement\:\:condition} \\[3ex] \underline{Multiplication\:\:Rule\:\:for\:\:Dependent\:\:Events} \\[3ex] red-blue \\[3ex] P(RB) = \dfrac{3}{12} * \dfrac{2}{11} \\[5ex] = \dfrac{1}{2} * \dfrac{1}{11} \\[5ex] = \dfrac{1 * 1}{2 * 11} \\[5ex] = \dfrac{1}{22} \\[5ex] blue-red \\[3ex] P(BR) = \dfrac{2}{12} * \dfrac{3}{11} \\[5ex] = \dfrac{1}{22} \\[5ex] red-blue\:\:OR\:\:blue-red \\[3ex] \underline{Addition\:\:Rule\:\:for\:\:Mutually-Exclusive\:\:Events} \\[3ex] P(RB\:\:OR\:\:BR) = \dfrac{1}{22} + \dfrac{1}{22} \\[5ex] = \dfrac{1 + 1}{22} \\[5ex] = \dfrac{2}{22} \\[5ex] = \dfrac{1}{11}$

(61.) WASSCE There are 5 more girls than boys in a class.
If 2 boys join the class, the ratio of girls to boys will be 5:4.
Find the probability of selecting a boy as the class prefect.

$\underline{Initial\:\:Count} \\[3ex] Let: \\[3ex] number\:\:of\:\:boys = p \\[3ex] number\:\:of\:\:girls = p + 5...(5\:\:more\:\:girls\:\:than\:\:boys) \\[3ex] \underline{Later\:\:Count...based\:\:on\:\:the\:\:condition,\:\: "if"} \\[3ex] number\:\:of\:\:boys = p + 2...(2\:\:more\:\:boys\:\:join) \\[3ex] number\:\:of\:\:girls = p + 5...(no\:\:change) \\[3ex] Ratio\:\:of\:\:girls:boys = 5:4 \\[3ex] \implies \dfrac{p + 5}{p + 2} = \dfrac{5}{4} \\[5ex] 4(p + 5) = 5(p + 2) \\[3ex] 4p + 20 = 5p + 10 \\[3ex] 20 - 10 = 5p - 4p \\[3ex] 10 = p \\[3ex] p = 10 \\[3ex] Number\:\:of\:\:boys = p = 10 \\[3ex] 10\:\:boys \\[3ex]$ Student: Excuse me Ma'am/Sir
I thought the boys would be $10 + 2 = 12$
Rather than $10$
Teacher: Yes, you have a point.
The initial number of boys is $p = 10$
However, the later count is based on a conditional statement, "if"
"If" $2$ boys join the class...then the ratio is ...
This does not imply that $2$ boys "actually" joined them
This was to assist us in determining the number of girls and boys in the class
So, we have to go by the Initial Count, rather than the "Conditional" Later Count.

$Number\:\:of\:\:girls \\[3ex] = p + 5 \\[3ex] = 10 + 5 = 15 \\[3ex] 15\:\:girls \\[3ex] total\:\:number\:\:of\:\:pupils\:\:in\:\:the\:\:class \\[3ex] 10 + 15 = 25 \\[3ex] 15\:\:pupils \\[3ex] n(S) = 25 \\[3ex] Let\:\:E\:\:be\:\:the\:\:event\:\:of\:\:selecting\:\:a\:\:boy\:\:as\:\:the\:\:class\:\:prefect \\[3ex] n(boys) = 10 \\[3ex] P(E) = \dfrac{n(boys)}{n(S)} \\[5ex] = \dfrac{10}{25} \\[5ex] = \dfrac{2}{5} \\[5ex]$ The probability of selecting a boy as the class prefect is $\dfrac{2}{5}$
(62.) WASSCE The probability that Mary will pass a class test is $\dfrac{3}{4}$
and the probability that Kofi was pass the same test is $\dfrac{3}{5}$

Find the probability that exactly one of them will pass the class test.

For exactly one of them to pass the test means that either:
Mary passes AND Kofi does not pass OR
Mary does not pass AND Kofi passes

$Let: \\[3ex] Mary\:\:passing\:\:the\:\:test = M \\[3ex] Kofi\:\:passing\:\:the\:\:test = K \\[3ex] P(M) = \dfrac{3}{4} \\[5ex] P(M') = 1 - \dfrac{3}{4} = \dfrac{4}{4} - \dfrac{3}{4} = \dfrac{1}{4} \\[5ex] P(K) = \dfrac{3}{5} \\[5ex] P(K') = 1 - \dfrac{3}{5} = \dfrac{5}{5} - \dfrac{3}{5} = \dfrac{2}{5} \\[5ex] P(exactly\:\:one\:\:passes) \\[3ex] = P(M) * P(K') + P(M') * P(K) \\[3ex] = \left(\dfrac{3}{4} * \dfrac{2}{5}\right) + \left(\dfrac{1}{4} * \dfrac{3}{5}\right) \\[5ex] = \left(\dfrac{3}{2} * \dfrac{1}{5}\right) + \dfrac{3}{20} \\[5ex] = \dfrac{3}{10} + \dfrac{3}{20} \\[5ex] LCD = 20 \\[3ex] = \dfrac{6}{20} + \dfrac{3}{20} \\[5ex] = \dfrac{6 + 3}{20} \\[5ex] = \dfrac{9}{20}$
(63.) WASSCE In a class, the probability that a student passes a test is $\dfrac{2}{5}$.

What is the probability that if 2 students are chosen at random from the class, one would pass and the other would fail?

$P(pass) = \dfrac{2}{5} \\[5ex] P(fail) = 1 - P(pass) ... Complementary\:\:Rule \\[3ex] P(fail) = 1 - \dfrac{2}{5} = \dfrac{5}{5} - \dfrac{2}{5} = \dfrac{5 - 2}{5} = \dfrac{3}{5} \\[5ex] P(one\:\:passes\:\:AND\:\:one\:\:fails) \\[3ex] = P[(1st\:\:passes\:\:AND\:\:2nd\:\:fails)\:\:OR\:\:(1st\:\:fails\:\:AND\:\:2nd\:\:passes)] \\[3ex] = P(1st\:\:passes\:\:AND\:\:2nd\:\:fails) + P(1st\:\:fails\:\:AND\:\:2nd\:\:passes) ... Addition\:\:Rule \\[3ex] P(1st\:\:passes\:\:AND\:\:2nd\:\:fails) = \dfrac{2}{5} * \dfrac{3}{5} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] P(1st\:\:fails\:\:AND\:\:2nd\:\:passes) = \dfrac{3}{5} * \dfrac{2}{5} ... Multiplication\:\:Rule - Independent\:\:Events \\[3ex] = \dfrac{2}{5} * \dfrac{3}{5} + \dfrac{3}{5} * \dfrac{2}{5} \\[5ex] = \dfrac{2 * 3}{5 * 5} + \dfrac{3 * 2}{5 * 5} \\[5ex] = \dfrac{6}{25} + \dfrac{6}{25} \\[5ex] = \dfrac{6 + 6}{25} \\[5ex] = \dfrac{12}{25}$
(64.) MB Exam A die and a 10 toea coin are tossed on a flat table together.
Results are then observed.
Find the probability of obtaining
(ii) a number greater than 4 and a head.
(iii) an even number and a tail.

Used: Punnett Square
Sample Space for the Toss of a Fair Die and a Fair Coin
A Die in the Column and A Coin in the Row
$1\:\:Coin\:\:\rightarrow$
$1\:\:Die\:\:\downarrow$
$H$ $T$
$1$ $1H$ $1T$
$2$ $2H$ $2T$
$3$ $3H$ $3T$
$4$ $4H$ $4T$
$5$ $5H$ $5T$
$6$ $6H$ $6T$

$S = \{ \\[3ex] 1H, 1T, \\[3ex] 2H, 2T, \\[3ex] 3H, 3T, \\[3ex] 4H, 4T, \\[3ex] 5H, 5T, \\[3ex] 6H, 6T \} \\[3ex] n(S) = 12 \\[5ex] (i.) \\[3ex] E = \{6H\} \\[3ex] n(E) = 1 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{1}{12} \\[5ex] (ii) \\[3ex] E =\{5H, 6H\} \\[3ex] n(E) = 2 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{2}{12} \\[5ex] P(E) = \dfrac{1}{6} \\[5ex] (iii) \\[3ex] E = \{2T, 4T, 6T\} \\[3ex] n(E) = 3 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{3}{12} \\[5ex] P(E) = \dfrac{1}{4}$
(65.) WASSCE A box contains five blue, three red and two white identical balls.
If 2 balls are selected at random, one after the other with replacement, from the box, find the probability of selecting:

(a.) two blue OR two red balls
(b.) one red AND one white ball.

$Let: \\[3ex] blue = B \\[3ex] n(B) = 5 \\[3ex] red = R \\[3ex] n(R) = 3 \\[3ex] white = W \\[3ex] n(W) = 2 \\[3ex] S = \{5B, 3R, 2W\} \\[3ex] n(S) = 5 + 3 + 2 = 10 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} \\[5ex] P(B) = \dfrac{5}{10} \\[5ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{3}{10} \\[5ex] (a.) \\[3ex] With\:\:Replacement\:\:condition \\[3ex] P(BB) = \dfrac{5}{10} * \dfrac{5}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(BB) = \dfrac{25}{100} \\[5ex] P(RR) = \dfrac{3}{10} * \dfrac{3}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(RR) = \dfrac{9}{100} \\[5ex] P(BB\:\:OR\:\:\:RR) = P(BB \cup RR) \\[3ex] P(BB \cup RR) = P(BB) + P(RR) - P(BB \cap RR) ...Addition\:\:Rule \\[3ex] P(BB \cap RR) = 0 ...Mutually\:\:Exclusive\:\:Events \\[3ex] \rightarrow P(BB \cup RR) = \dfrac{25}{100} + \dfrac{9}{100} \\[5ex] P(BB \cup RR) = \dfrac{25 + 9}{100} \\[5ex] P(BB \cup RR) = \dfrac{34}{100} \\[5ex] P(BB \cup RR) = \dfrac{17}{50} \\[5ex] (b) \\[3ex] 1\:\:Red\:\:AND\:\:1\:\:White \implies in\:\:any\:\:order \\[3ex] This\:\:means\:\:RW\:\:OR\:\:WR \\[3ex] P(RW) = \dfrac{3}{10} * \dfrac{2}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(RW) = \dfrac{6}{100} \\[5ex] P(WR) = \dfrac{2}{10} * \dfrac{3}{10} ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[5ex] P(WR) = \dfrac{6}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{6}{100} + \dfrac{6}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{6 + 6}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{12}{100} \\[5ex] P(RW\:\:OR\:\:WR) = \dfrac{3}{25}$
(66.) GCSE A bus can be early, on time, or late.
The probability that the bus is early is 0.1
The probability that the bus is on time is 0.6
Work out the probability that the bus is late.

Because the bus can be early, on time, or late (3 conditions),
The probability that the bus is early plus the probability that the bus is on time plus the probability that the bus is late is $1$

$P(early) + P(on\;\;time) + P(late) = 1 \\[3ex] 0.1 + 0.6 + P(late) = 1 \\[3ex] 0.7 + P(late) = 1 \\[3ex] P(late) = 1 - 0.7 \\[3ex] P(late) = 0.3 \\[3ex]$ The probability that the bus is late is $0.3$
(67.) ACT Bella will pick 1 jelly bean at random out of a bag containing 28 jelly beans that are in the colors and quantities shown in the table below.
Each of the jelly beans is 1 color only.

 Color Quantity Green Black Red Orange Yellow Blue $6$ $3$ $5$ $2$ $4$ $8$

What is the probability that Bella will pick a blue or yellow jelly bean?

Let the event of picking a blue jelly bean be $B$
and the event of picking a yellow jelly bean be $Y$

$n(S) = 28 \\[3ex] n(B) = 8 \\[3ex] n(Y) = 4 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} = \dfrac{8}{28} \\[5ex] P(Y) = \dfrac{n(Y)}{n(S)} = \dfrac{4}{28} \\[5ex] P(B\:\:\:AND\:\:\:Y) = 0 ...Mutually\:\:Exclusive\:\:Event \\[3ex] P(B\:\:\:OR\:\:\:Y) = P(B) + P(Y) - P(B\:\:\:AND\:\:\:Y) \\[3ex] = \dfrac{8}{28} + \dfrac{4}{28} - 0 \\[5ex] = \dfrac{8 + 4}{28} \\[5ex] = \dfrac{12}{28} \\[5ex] P(B\:\:\:OR\:\:\:Y) = \dfrac{3}{7}$
(68.) ACT One morning at a coffee shop, each customer ordered either decaf or regular coffee, and each ordered it either with milk or without milk.
The number of customers who ordered each type of coffee with or without milk is listed in the table below.

 Order Decaf Regular Total With milk Without milk $12$ $6$ $8$ $10$ $20$ $16$ Total $18$ $18$ $36$

A customer will be randomly selected from all $36$ customers for a prize.
What is the probability that the selected customer will have ordered a regular coffee without milk?

$A.\:\: \dfrac{1}{6} \\[5ex] B.\:\: \dfrac{5}{18} \\[5ex] C.\:\: \dfrac{5}{13} \\[5ex] D.\:\: \dfrac{1}{2} \\[5ex] E.\:\: \dfrac{5}{8} \\[5ex]$

Let $E$ be the event that the selected customer ordered a regular coffee without milk.

$n(S) = 36 \\[3ex] n(E) = 10 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{10}{36} \\[5ex] = \dfrac{5}{18}$
(69.) ACT A rectangle, with its vertex coordinates labeled, is graphed in the standard (x, y) coordinate plane below.
A lattice point is a point with coordinates that are both integers.
A lattice point inside but NOT on the rectangle will be chosen at random.
What is the probability that the sum of the $x-coordinate$ and the $y-coordinate$ of the chosen lattice point will be odd?

NOTE:
A lattice point is a point with coordinates that are both integers
We are concerned with the lattice points only inside the rectangle

For the $x-coordinate$, the integers inside the rectangle are $0$ through $6$ (both excluded) = $1, 2, 3, 4, 5$

For the $y-coordinate$, the integers inside the rectangle are $0$ through $4$ (both excluded) = $1, 2, 3$

 $y\downarrow\:\:(+)\:\:x\rightarrow$ $1$ $2$ $3$ $4$ $5$ $1$ $2$ $\color{red}{3}$ $4$ $\color{red}{5}$ $6$ $2$ $\color{red}{3}$ $4$ $\color{red}{5}$ $6$ $\color{red}{7}$ $3$ $4$ $\color{red}{5}$ $6$ $\color{red}{7}$ $8$

Numbers in red are odd numbers
Let $E$ be the event of selecting odd numbers

$n(E) = 7 \\[3ex] n(S) = 3 * 5 = 15 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{7}{15}$
(70.) ACT The 35-member History Club is meeting to choose a student government representative.
The members decide that the representative, who will be chosen at random, CANNOT be any of the 3 officers of the club.
What is the probability that Hiroko, who is a member of the club but NOT an officer, will be chosen?

$F.\:\: 0 \\[3ex] G.\:\: \dfrac{4}{35} \\[5ex] H.\:\: \dfrac{1}{35} \\[5ex] J.\:\: \dfrac{1}{3} \\[5ex] K.\:\: \dfrac{1}{32} \\[5ex]$ Before you solve this question, check out Questions (13.) and (16.)
Did you notice anything?

$Let\:\: Member = M \\[3ex] Let\:\:Officer = F \\[3ex] n(S) = 35 \\[3ex] n(M) = 32 \\[3ex] n(F) = 3 \\[3ex] n(Hiroko) = 1 \\[3ex] Hiroko\:\: \epsilon \:\:M \\[3ex] P(Hiroko) = \dfrac{n(Hiroko)}{n(M)} \\[5ex] P(Hiroko) = \dfrac{1}{32}$
(71.) NSC Veli and Bongi are learners at the same school.
Some days they arrive late at school.
The probability that neither Veli nor Bongi will arrive late on any day is 0.7

(71.1.1) Calculate the probability that at least one of the two learners will arrive late on a randomly selected day.

(71.1.2) The probability that Veli arrives late for school on a randomly selected day is 0.25, while the probability that both of them arrive late for school on that day is 0.15
Calculate the probability that Bongi will arrive late for school on that day.

(71.1.3) The principal suspects that the latecoming of the two learners is linked.
The principal asks you to determine whether the events of Veli arriving late for school and Bongi arriving late for school are statistically independent or not.
What will be your response to him?
Show ALL calculations.

Let the:
Event that Veli will arrive late at school = $V$
This implies that the event that Veli will not arrive late at school = $V'$

Event that Bongi will arrive late at school = $B$
This implies that the event that Bongi will not arrive late at school = $B'$

Event that both Veli and Bongi will arrive late at school = $VB$
This implies that the event that neither Veli nor Bongi will arrive late at school = $B'V'$
This also means that the event that both Veli and Bongi will not arrive at school = $B'V'$

$P(V'B') = 0.7 \\[3ex] (71.1.1) \\[3ex] P(at\:\:least\:\:one\:\:will) + P(none\:\:will) = 1...Complementary\:\:Rule \\[3ex] P(at\:\:least\:\:one\:\:will) + P(V'B') = 1 \\[3ex] P(at\:\:least\:\:one\:\:will) = 1 - P(V'B') \\[3ex] P(at\:\:least\:\:one\:\:will) = 1 - 0.7 \\[3ex] P(at\:\:least\:\:one\:\:will) = 0.3 \\[3ex] (71.1.2) \\[3ex]$
(72.) ACT A wallet containing 5 five-dollar bills, 7 ten-dollar bills, and 8 twenty-dollar bills is found and returned to its owner.
The wallet's owner will reward the finder with 1 bill drawn randomly from the wallet.
What is the probability that the bill drawn will be a twenty-dollar bill?

$A.\:\: \dfrac{1}{20} \\[5ex] B.\:\: \dfrac{4}{51} \\[5ex] C.\:\: \dfrac{1}{8} \\[5ex] D.\:\: \dfrac{2}{5} \\[5ex] E.\:\: \dfrac{2}{3} \\[5ex]$

$Let: \\[3ex] five-dollar\:\:bill = F \\[3ex] n(F) = 5 \\[3ex] ten-dollar\:\:bill = T \\[3ex] n(T) = 7 \\[3ex] twenty-dollar\:\:bill = W \\[3ex] n(W) = 8 \\[3ex] S = \{5F, 7T, 8W\} \\[3ex] n(S) = 5 + 7 + 8 = 20 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] P(W) = \dfrac{8}{20} \\[5ex] P(W) = \dfrac{2}{5}$
(73.) A coin is flipped four times.
For each of the sets described below, express the event as a set in roster notation.
Each outcome is written as a string of length 4 from {H, T} such as {HHHT}.
Assuming the coin is a fair coin, give the probability of each event.

(a.) The first and last flips come up heads.

(b.) There are at least two consecutive flips that come up heads.

(c.) The first flip comes up tails and there are at least two consecutive flips that come up heads.

Used: Punnett Square
Sample Space for the Four Flips of a Fair Coin
$2$ flips in the Column and $2$ flips in the Row
$2\:\:Flips\:\:\rightarrow$
$2\:\:Flips\:\:\downarrow$
$HH$ $HT$ $TH$ $TT$
$HH$ $HHHH$ $HHHT$ $HHTH$ $HHTT$
$HT$ $HTHH$ $HTHT$ $HTTH$ $HTTT$
$TH$ $THHH$ $THHT$ $THTH$ $THTT$
$TT$ $TTHH$ $TTHT$ $TTTH$ $TTTT$

$S = \{ \\[3ex] HHHH, HHHT, HHTH, HHTT, \\[3ex] HTHH, HTHT, HTTH, HTTT, \\[3ex] THHH, THHT, THTH, THTT, \\[3ex] TTHH, TTHT, TTTH, TTTT \\[3ex] \} \\[3ex] n(S) = 16 \\[5ex] (a.) \\[3ex] E = \{HHHH, HHTH, HTHH, HTTH\} \\[3ex] n(E) = 4 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{4}{16} \\[5ex] P(E) = \dfrac{1}{4} \\[5ex]$ For the second question:
At least two means two or more
At least two consecutive means: first and second or second and third or third and fourth or
first, second, and third or second, third, and fourth
At least two consecutive flips that come up heads means that:
first and second is $HH$ or second and third is $HH$ or or third and fourth is $HH$ or
first, second, and third is $HH$ or second, third, and fourth is $HH$

$(b.) \\[3ex] E = \{HHHH, HHHT, HHTH, HHTT, HTHH, THHH, THHT, TTHH\} \\[3ex] n(E) = 8 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{8}{16} \\[5ex] P(E) = \dfrac{1}{2} \\[5ex]$ For the third question:
The second question must be satisfied in addition to the requirement that the first flip is a $T$
So, let us look at the event space in the answer to the second question and select only the events where the first flip results in a tail.

$(c.) \\[3ex] E = \{THHH, THHT, TTHH\} \\[3ex] P(E) = \dfrac{3}{16}$
(74.) JAMB Some white balls were put in a basket containing twelve red balls and sixteen black balls.
If the probability of picking a white ball from the basket is $\dfrac{3}{7}$, how many white balls were introduced?

$A.\:\: 32 \\[3ex] B.\:\: 28 \\[3ex] C.\:\: 21 \\[3ex] D.\:\: 12 \\[3ex]$

We can solve this question in two ways.
Use any method you prefer.

$Let: \\[3ex] white = W \\[3ex] red = R \\[3ex] black = B \\[3ex] \underline{First\:\:Method} \\[3ex] n(W) = p \\[3ex] n(R) = 12 \\[3ex] n(B) = 16 \\[3ex] n(S) = p + 12 + 16 = p + 28 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] P(W) = \dfrac{p}{p + 28} \\[5ex] P(W) = \dfrac{3}{7} \\[5ex] \rightarrow \dfrac{p}{p + 28} = \dfrac{3}{7} \\[5ex] Cross\:\:multiply \\[3ex] 7p = 3(p + 28) \\[3ex] 7p = 3p + 84 \\[3ex] 7p - 3p = 84 \\[3ex] 4p = 84 \\[3ex] p = \dfrac{84}{4} \\[5ex] p = 21 \\[3ex] \underline{Second\:\:Method} \\[3ex] P(W) = \dfrac{3}{7} \\[5ex] P(W) + P(W') = 1 ...Complementary\:\:Rule \\[3ex] P(W') = 1 - P(W) \\[3ex] P(W') = 1 - \dfrac{3}{4} \\[5ex] P(W') = \dfrac{7}{7} - \dfrac{3}{7} \\[5ex] P(W') = \dfrac{7 - 3}{7} \\[5ex] P(W') = \dfrac{4}{7} \\[5ex] n(W') = n(R) + n(B) \\[3ex] n(W') = 12 + 16 = 28 \\[3ex] P(W') = \dfrac{n(W')}{n(S)} \\[5ex] n(S) * P(W') = n(W') \\[3ex] n(S) = \dfrac{n(W')}{P(W')} \\[5ex] n(S) = n(W') \div P(W') \\[3ex] n(S) = 28 \div \dfrac{4}{7} \\[5ex] n(S) = 28 * \dfrac{7}{4} \\[5ex] n(S) = 7(7) \\[3ex] n(S) = 49 \\[3ex] Additional\:\:white\:\:balls = 49 - 28 \\[3ex] Additional\:\:white\:\:balls = 21 \\[3ex]$ $21$ white balls were introduced in the basket.
(75.) MB Exam What is the probability of getting:
(i) a red ten (10) from a pack of cards.
(ii) 2 boys born to a couple.
(iii) A sum of 6 on two dice thrown together

Note: Write All answers as simplest fractions.

(i)
There are two red $10$ in a standard deck of cards: the Diamond $10$ and the Heart $10$

$n(S) = 52 \\[3ex] E = \{Diamond-10, Heart-10\} \\[3ex] n(E) = 2 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{2}{52} \\[5ex] P(E) = \dfrac{1}{26} \\[5ex] (ii) \\[3ex] Let: \\[3ex] Boy = B \\[3ex] Girl = G \\[3ex] P(B) = \dfrac{1}{2} \\[5ex] P(G) = \dfrac{1}{2}...will\;\;not\;\;be\;\;used \\[5ex] Because\;\; \\[3ex] 1st\;\;child\;\;is\;\;a\;\;boy\;\;AND\;\;2nd\;\;child\;\;is\;\;a\;\;boy \\[3ex] P(BB) = \dfrac{1}{2} * \dfrac{1}{2} ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[5ex] P(BB) = \dfrac{1 * 1}{2 * 2} \\[5ex] P(BB) = \dfrac{1}{4} \\[5ex] (iii) \\[3ex]$ Used: Punnett Square
Sample Space for a Toss of Two Fair Dice
A Die in the Column and A Die in the Row
$1\:\:Die\:\:\rightarrow$
($+$)
$1\:\:Die\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $2$ $3$ $4$ $5$ $6$ $7$
$2$ $3$ $4$ $5$ $6$ $7$ $8$
$3$ $4$ $5$ $6$ $7$ $8$ $9$
$4$ $5$ $6$ $7$ $8$ $9$ $10$
$5$ $6$ $7$ $8$ $9$ $10$ $11$
$6$ $7$ $8$ $9$ $10$ $11$ $12$

$n(S) = 36 \\[5ex] E = \{6, 6, 6, 6, 6\} \\[3ex] n(E) = 5 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{5}{36}$
(76.) WASSCE Three red balls, five green balls and a number of blue balls are put together in a sack.
One ball is picked at random from the sack.
If the probability of picking a red ball is $\dfrac{1}{6}$, find:
(a) the number of blue balls in the sack.
(b) the probability of picking a green ball.

$Let: \\[3ex] red = R \\[3ex] green = G \\[3ex] blue = B \\[3ex] n(R) = 3 \\[3ex] n(G) = 5 \\[3ex] n(B) = x \\[3ex] n(S) = 3 + 5 + x \\[3ex] n(S) = 8 + x \\[3ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] P(R) = \dfrac{3}{8 + x} \\[5ex] Also:\;\;P(R) = \dfrac{1}{6} \\[5ex] (a) \\[3ex] P(R) = P(R) \\[3ex] \implies \dfrac{3}{8 + x} = \dfrac{1}{6} \\[5ex] Cross\;\;Multiply \\[3ex] (8 + x) * 1 = 3 * 6 \\[3ex] 8 + x = 18 \\[3ex] x = 18 - 8 \\[3ex] x = 10 \\[3ex]$ There are ten blue balls in the sack.

$(b) \\[3ex] n(S) = 8 + x \\[3ex] n(S) = 8 + 10 \\[3ex] n(S) = 18 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} \\[5ex] P(G) = \dfrac{5}{18} \\[5ex]$ The probability of picking a green ball is $\dfrac{5}{18}$
(77.) WASSCE: FM Two fair dice are thrown together two times.
Find the probability of obtaining a sum of seven in the first throw and a sum of four in the second throw.

There are at least two options of solving this question.

One option that comes to mind:
Two fair coins tossed two times? (Not two fair coins tossed one time)...
Drawing the Punnett Square or Tree Diagram for this case would be cumbersome.
Yes, it is cumbersome.
So, why not do it this way?
The events are independent.
Look at the keyword in the question: AND
So, let us use the Multiplication Rule for Independent Events
We shall consider these events:
Event One: Two fair dice tossed one time (first throw): selecting a sum of seven
Event Two: Two fair dice tossed one time (second throw): selecting a sum of four
This option is recommended.

Used: Punnett Square
Sample Space for a Toss of Two Fair Dice (First Throw)
A Die in the Column and A Die in the Row
$1\:\:Die\:\:\rightarrow$
($+$)
$1\:\:Die\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $2$ $3$ $4$ $5$ $6$ $7$
$2$ $3$ $4$ $5$ $6$ $7$ $8$
$3$ $4$ $5$ $6$ $7$ $8$ $9$
$4$ $5$ $6$ $7$ $8$ $9$ $10$
$5$ $6$ $7$ $8$ $9$ $10$ $11$
$6$ $7$ $8$ $9$ $10$ $11$ $12$

$n(S) = 36 \\[5ex] E_1 = \{7, 7, 7, 7, 7, 7\} \\[3ex] n(E_1) = 6 \\[3ex] P(E_1) = \dfrac{n(E_1)}{n(S)} \\[5ex] P(E_1) = \dfrac{6}{36} \\[5ex] P(E_1) = \dfrac{1}{6} \\[5ex] \implies P(sum\;\;of\;\;7\;\;first\;\;throw) = \dfrac{1}{6} \\[5ex]$ Sample Space for a Toss of Two Fair Dice (Second Throw)
A Die in the Column and A Die in the Row
$1\:\:Die\:\:\rightarrow$
($+$)
$1\:\:Die\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $2$ $3$ $4$ $5$ $6$ $7$
$2$ $3$ $4$ $5$ $6$ $7$ $8$
$3$ $4$ $5$ $6$ $7$ $8$ $9$
$4$ $5$ $6$ $7$ $8$ $9$ $10$
$5$ $6$ $7$ $8$ $9$ $10$ $11$
$6$ $7$ $8$ $9$ $10$ $11$ $12$

$n(S) = 36 \\[5ex] E_2 = \{4, 4, 4\} \\[3ex] n(E_2) = 3 \\[3ex] P(E_2) = \dfrac{n(E)}{n(S)} \\[5ex] P(E_2) = \dfrac{3}{36} \\[5ex] P(E_2) = \dfrac{1}{12} \\[5ex] \implies P(sum\;\;of\;\;4\;\;second\;\;throw) = \dfrac{1}{12} \\[5ex] therefore \;\; P(sum\;\;of\;\;7\;\;first\;\;throw) \;\;AND\;\; P(sum\;\;of\;\;4\;\;second\;\;throw) \\[5ex] = \dfrac{1}{6} * \dfrac{1}{12} ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[5ex] = \dfrac{1 * 1}{6 * 12} \\[5ex] = \dfrac{1}{72}$
(78.) WASSCE: FM Three soldiers, X, Y and Z have probabilities $\dfrac{1}{3}, \dfrac{1}{5}$ and $\dfrac{1}{4}$ respectively of hitting a target.
If each of them fires once, find, correct to two decimal places, the probability that only one of them hits the target.

If only one of them hits the target, this means that:
1st Case: $X$ hits it AND $Y$ misses it AND $Z$ misses it
OR
2nd Case: $X$ misses it AND $Y$ hits it AND $Z$ misses it
OR
3rd Case: $X$ misses it AND $Y$ misses it AND $Z$ hits it
So, we have these three cases.

$P(X) = \dfrac{1}{3}...hits\;\;it \\[5ex] P(X') = 1 - \dfrac{1}{3} = \dfrac{2}{3}...Complementary\;\;Rule...misses\;\;it \\[5ex] P(Y) = \dfrac{1}{5}...hits\;\;it \\[5ex] P(Y') = 1 - \dfrac{1}{5} = \dfrac{4}{5}...Complementary\;\;Rule...misses\;\;it \\[5ex] P(Z) = \dfrac{1}{4}...hits\;\;it \\[5ex] P(Z') = 1 - \dfrac{1}{4} = \dfrac{3}{4}...Complementary\;\;Rule...misses\;\;it \\[5ex] 1st\;\;Case \\[3ex] P(X) * P(Y') * P(Z') \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{1}{3} * \dfrac{4}{5} * \dfrac{3}{4} \\[5ex] = \dfrac{1}{5} \\[5ex] 2nd\;\;Case \\[3ex] P(X') * P(Y) * P(Z') \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{2}{3} * \dfrac{1}{5} * \dfrac{3}{4} \\[5ex] = \dfrac{1}{5} * \dfrac{1}{2} \\[5ex] = \dfrac{1}{10} \\[5ex] 3rd\;\;Case \\[3ex] P(X') * P(Y') * P(Z) \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{2}{3} * \dfrac{4}{5} * \dfrac{1}{4} \\[5ex] = \dfrac{2}{3} * \dfrac{1}{5} \\[5ex] = \dfrac{2}{15} \\[5ex] P(only\;\;one\;\;hits\;\;the\;\;target) \\[3ex] = 1st\;\;Case + 2nd\;\;Case + 3rd\;\;Case \\[3ex] ... Addition\;\;Rule\;\;for\;\;Mutually\;\;Exclusive\;\;Events \\[3ex] Mutually\;\;exclusive\;\;because\;\;only\;\;one\;\;hits\;\;it \\[3ex] Only\;\;one\;\;event\;\;can\;\;occur\;\;at\;\;a\;\;time \\[3ex] Two\;\;or\;\;more\;\;cases\;\;cannot\;\;occur\;\;at\;\;the\;\;same\;\;time \\[3ex] = \dfrac{1}{5} + \dfrac{1}{10} + \dfrac{2}{15} \\[5ex] = \dfrac{6 + 3 + 4}{30} \\[5ex] = \dfrac{13}{30} \\[5ex] = 0.4333333333 \\[3ex] \approx 0.43...to\;\;two\;\;decimal\;\;places \\[3ex]$ The probability that only one of them hits the target is approximately $0.43$
(79.) GCSE A number is picked at random from the first four prime numbers.
A number is picked at random from the first four square numbers.
The two numbers are added to get a score.

(a.) Complete the table.

(b.) What is the probability that the score is a prime number?

(a.)
Used: Punnett Square
Sample Space for the First Four Prime Numbers and the First Four Square Numbers
$Square\;\;numbers\:\:\rightarrow$
($+$)
$Prime\;\;numbers\:\:\downarrow$
$1$ $4$ $9$ $16$
$2$ $3$ $6$ $11$ $18$
$3$ $4$ $7$ $12$ $19$
$5$ $6$ $9$ $14$ $21$
$7$ $8$ $11$ $16$ $23$

$n(S) = 4 * 4 = 16 \\[5ex] (b.) \\[3ex] Let\;\;E\;\;be\;\;the\;\;event\;\;the\;\;score\;\;a\;\;prime\;\;number \\[3ex] E = \{3, 7, 11, 11, 19, 23\} \\[3ex] n(E) = 6 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E) = \dfrac{6}{16} \\[5ex] P(E) = \dfrac{3}{8} \\[5ex]$ The probability that the score is a prime number is $\dfrac{3}{8}$
(80.) JAMB Teams P and Q are involved in a game of football.
What is the probability that the game ends in a draw?

$A.\:\: \dfrac{2}{3} \\[5ex] B.\:\: \dfrac{1}{2} \\[5ex] C.\:\: \dfrac{1}{3} \\[5ex] D.\:\: \dfrac{1}{4} \\[5ex]$

Some will argue that this is an incomplete question
However, we need to solve this because it is a question that was asked.
Besides, none of the options is: Insufficient information to solve the question.

In analyzing this question:
The entire possibilities are:
1st Case: Team $P$ wins AND Team $Q$ loses OR
2nd Case: Team $P$ loses AND Team $Q$ wins OR
3rd Case: Team $P$ and Team $Q$ game ends in a draw

Team $P$ and Team $Q$ has an equal probability of winning the game.
Let $P$ be the event that Team $P$ wins the game.
$P'$ is the event that Team $P$ loses.
Let $Q$ be the event that Team $Q$ wins the game.
$Q'$ is the event that Team $Q$ loses.

$P(P) = 50\% = \dfrac{1}{2} \\[5ex] P(Q) = 50\% = \dfrac{1}{2} \\[5ex] P(P') = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\[5ex] P(Q') = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\[5ex] \underline{1st\;\;Case} \\[3ex] P(PQ') = \dfrac{1}{2} * \dfrac{1}{2} \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] P(PQ') = \dfrac{1}{4} \\[5ex] \underline{2nd\;\;Case} \\[3ex] P(P'Q) = \dfrac{1}{2} * \dfrac{1}{2} \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] P(P'Q) = \dfrac{1}{4} \\[5ex] 1st\;\;Case + 2nd\;\;Case + 3rd\;\;Case = 1 \\[3ex] ...Sum\;\;of\;\;all\;\;probabilities \\[3ex] 3rd\;\;Case = P(Draw) \\[3ex] \dfrac{1}{4} + \dfrac{1}{4} + P(Draw) = 1 \\[5ex] \dfrac{1 + 1}{4} + P(Draw) = 1 \\[5ex] \dfrac{2}{4} + P(Draw) = \dfrac{4}{4} \\[5ex] P(Draw) = \dfrac{4}{4} - \dfrac{2}{4} \\[5ex] P(Draw) = \dfrac{4 - 2}{4} \\[5ex] P(Draw) = \dfrac{2}{4} \\[5ex] P(Draw) = \dfrac{1}{2}$

(81.) HSC Mathematics Standard 2 The top of a rectangular table is divided into 8 equal sections as shown.
 1 2 3 4 5 6 7 8

A standard die with faces labelled 1 to 6 is rolled onto the table.
The die is equally likely to land in any of the 8 sections of the table.
If the die does not land entirely in one section of the table, it is rolled again.
A score is calculated by multiplying the value shown on the top face of the die by the number shown in the section of the table where the die lands.
What is the probability of getting a score of 6?

$A.\;\; \dfrac{1}{48} \\[5ex] B.\;\; \dfrac{1}{12} \\[5ex] C.\;\; \dfrac{1}{8} \\[5ex] D.\;\; \dfrac{1}{6} \\[5ex]$

The product, 6 is marked in red.
$Rectangular\;\;Table\:\:\rightarrow \\[1ex] \;\;\;\;\;* \\[1ex] Standard\;\;Die\:\:\downarrow$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$1$ $1$ $2$ $3$ $4$ $5$ $\color{red}6$ $7$ $8$
$2$ $2$ $4$ $\color{red}6$ $8$ $10$ $12$ $14$ $16$
$3$ $3$ $\color{red}6$ $9$ $12$ $15$ $18$ $21$ $24$
$4$ $4$ $8$ $12$ $16$ $20$ $24$ $28$ $32$
$5$ $5$ $10$ $15$ $20$ $25$ $30$ $35$ $40$
$6$ $\color{red}6$ $12$ $18$ $24$ $30$ $36$ $42$ $48$

$n(S) = 6 * 8 = 48 \\[3ex] n(6) = 4 \\[3ex] P(6) = \dfrac{n(6)}{n(S)} \\[5ex] = \dfrac{4}{48} \\[5ex] = \dfrac{1}{12}$
(82.) ACT Sets A, B, and C are defined below.
A = {1, 2, 3, 4, 5, 6}
B = {2, 4, 6}
C= {1, 2}
A number will be randomly selected from set A.
What is the probability that the selected number will be an element of set B and an element of set C?

$A.\;\; 0 \\[3ex] B.\;\; \dfrac{1}{6} \\[5ex] C.\;\; \dfrac{2}{6} \\[5ex] D.\;\; \dfrac{4}{6} \\[5ex] E.\;\; \dfrac{5}{6} \\[5ex]$

$A = \{1, 2, 3, 4, 5, 6\} \\[3ex] n(A) = 6 \\[3ex] B \cap C = \{2\} \\[3ex] n(B \cap C) = 1 \\[3ex] P(B \cap C) \\[3ex] = \dfrac{n(B \cap C)}{n(A)} \\[5ex] = \dfrac{1}{6}$
(83.) WASSCE:FM A fair die with faces 1, 2, 3, 4, 5 and 6 is tossed twice.
Calculate the probability that the sum of numbers that showed up is:
(a.) multiple of 3
(b.) between 3 and 6

Sample Space for Two Tosses of A Fair Die
A Toss in the Column and a Toss in the Row
$1\:\:Toss\:\:\rightarrow$
($+$)
$1\:\:Toss\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $2$ $\color{red}{3}$ $\color{violet}{4}$ $\color{violet}{5}$ $\color{red}{6}$ $7$
$2$ $\color{red}{3}$ $\color{violet}{4}$ $\color{violet}{5}$ $\color{red}{6}$ $7$ $8$
$3$ $\color{violet}{4}$ $\color{violet}{5}$ $\color{red}{6}$ $7$ $8$ $\color{red}{9}$
$4$ $\color{violet}{5}$ $\color{red}{6}$ $7$ $8$ $\color{red}{9}$ $10$
$5$ $\color{red}{6}$ $7$ $8$ $\color{red}{9}$ $10$ $11$
$6$ $7$ $8$ $\color{red}{9}$ $10$ $11$ $\color{red}{12}$

(a.)
Multiples of 3 are marked in red

$n(S) = 6 * 6 = 36 \\[3ex] Let\;\;E = event\;\;of\;\;multiple\;\;of\;\;3 \\[3ex] n(E) = 12 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{12}{36} = \dfrac{1}{3} \\[5ex]$ (b.)
Sum of the numbers between 3 and 6 are marked in violet

$Let\;\;E = event\;\;of\;\;numbers\;\;between\;\;3\;\;and\;\;6 \\[3ex] n(E) = 7 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{7}{36}$
(84.) HSC Mathematics Standard 2 A bowl of fruit contains 17 apples of which 9 are red and 8 are green.
Dennis takes one apple at random and eats it.
Margaret also takes an apple at random and eats it.
By drawing a probability tree diagram, or otherwise, find the probability that Dennis and Margaret eat apples of the same colour.

(a.) Eating two apples of the same colour implies two red apples OR two green apples
This means that:
The first apple is red AND the second apple is red
OR
The first apple is green AND the second apple is green

(b.) Eating two apples, one at a time implies without replacement condition

$Let: \\[3ex] red = R \\[3ex] green = G \\[3ex] n(R) = 9 \\[3ex] n(G) = 8 \\[3ex] n(S) = 17 \\[3ex] P(RR)\;\;OR\;\;P(GG) \\[3ex] = \left(\dfrac{9}{17} * \dfrac{8}{16}\right) + \left(\dfrac{8}{17} * \dfrac{7}{16}\right) \\[5ex] = \dfrac{9}{34} + \dfrac{7}{34} \\[5ex] = \dfrac{16}{54} \\[5ex] = \dfrac{8}{17}$
(85.) ATAR: Mathematics Methods A charity organisation has printed 'Lucky 7' scratchie tickets as a fundraiser for use at two special events.
The tickets contain two panels.
Each ticket has the same numbers as the sample ticket shown below, arranged randomly and hidden within each panel.

A player scratches one section of each panel to reveal a number.
The two numbers revealed are then added together.
If the total is seven or higher, the player wins a prize.
At the first event, 400 tickets are purchased, and a prize is won on 124 occasions.
Let p denote the probability that a prize is won.
(a.) Determine the sample proportion of times that a prize is won at the first event.
(b.) Show that the probability p of winning a prize is $\dfrac{7}{24}$

$(a.) \\[3ex] sample\;\;proportion = \dfrac{number\;\;of\;\;prizes}{number\;\;of\;\;tickets} = \dfrac{124}{400} \\[5ex] = \dfrac{31}{100} \\[5ex]$ (b.)
Winning a prize implies having a sum of at least 7
The sum of at least 7 (7 or higher) is marked in red

Sample Space for the two Panels
Panel 1 in the Column and Panel 2 in the Row
$Panel\;2\:\:\rightarrow$
($+$)
$Panel\:1\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $2$ $3$ $4$ $5$ $6$ $\color{red}{7}$
$2$ $3$ $4$ $5$ $6$ $\color{red}{7}$ $\color{red}{8}$
$3$ $4$ $5$ $6$ $\color{red}{7}$ $\color{red}{8}$ $\color{red}{9}$
$1$ $2$ $3$ $4$ $5$ $6$ $\color{red}{7}$
$1$ $2$ $3$ $4$ $5$ $6$ $\color{red}{7}$
$1$ $2$ $3$ $4$ $5$ $6$ $\color{red}{7}$
$3$ $4$ $5$ $6$ $\color{red}{7}$ $\color{red}{8}$ $\color{red}{9}$
$2$ $3$ $4$ $5$ $6$ $\color{red}{7}$ $\color{red}{8}$

$n(S) = 8 * 6 = 48 \\[3ex] n(\ge 7) = 14 \\[3ex] P(winning\;\;a\;\;prize) = P(\ge 7) \\[3ex] = \dfrac{n(\ge 7)}{n(S)} \\[5ex] = \dfrac{14}{48} \\[5ex] = \dfrac{7}{24}$
(86.) ACT The numbers 1 through 15 were each written on individual pieces of paper, 1 number per piece.
Then the 15 pieces of paper were put in a jar.
One piece of paper will be drawn from the jar at random.
What is the probability of drawing a piece of paper with a number less than 9 written on it?

$A.\;\; \dfrac{1}{9} \\[5ex] B.\;\; \dfrac{1}{15} \\[5ex] C.\;\; \dfrac{6}{15} \\[5ex] D.\;\; \dfrac{7}{15} \\[5ex] E.\;\; \dfrac{8}{15} \\[5ex]$

$S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15\} \\[3ex] n(S) = 15 \\[3ex] E = \{1, 2, 3, 4, 5, 6, 7, 8\} \\[3ex] n(E) = 8 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{8}{15}$
(87.) ACT An 8-sided game piece with faces numbered from 1 to 8 is rolled twice.
What is the probability that an 8 is rooled both times?
(Note: Assume that each side has an equally likely chance of being rolled.)

$F.\;\; \dfrac{1}{64} \\[5ex] G.\;\; \dfrac{1}{32} \\[5ex] H.\;\; \dfrac{1}{8} \\[5ex] J.\;\; \dfrac{1}{4} \\[5ex] K.\;\; \dfrac{1}{2} \\[5ex]$

8-sided game: First Roll in the Column and Second Roll in the Row
$2nd\:\:Roll\:\:\rightarrow$
($+$)
$1st\:\:Roll\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$1$ $1, 1$ $1, 2$ $1, 3$ $1, 4$ $1, 5$ $1, 6$ $1, 7$ $1, 8$
$2$ $2, 1$ $2, 2$ $2, 3$ $2, 4$ $2, 5$ $2, 6$ $2, 7$ $2, 8$
$3$ $3, 1$ $3, 2$ $3, 3$ $3, 4$ $3, 5$ $3, 6$ $3, 7$ $3, 8$
$4$ $4, 1$ $4, 2$ $4, 3$ $4, 4$ $4, 5$ $4, 6$ $4, 7$ $4, 8$
$5$ $5, 1$ $5, 2$ $5, 3$ $5, 4$ $5, 5$ $5, 6$ $5, 7$ $5, 8$
$6$ $6, 1$ $6, 2$ $6, 3$ $6, 4$ $6, 5$ $6, 6$ $6, 7$ $6, 8$
$7$ $7, 1$ $7, 2$ $7, 3$ $7, 4$ $7, 5$ $7, 6$ $7, 7$ $7, 8$
$8$ $8, 1$ $8, 2$ $8, 3$ $8, 4$ $8, 5$ $8, 6$ $8, 7$ $8, 8$

$n(S) = 8 * 8 = 64 \\[3ex] E = \{8, 8\} \\[3ex] n(E) = 1 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[3ex] = \dfrac{1}{64}$
(88.) GCSE $\mathcal{E}$ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 5, 6, 8, 9}
B = {2, 6, 9}

(a) Complete the Venn diagram to represent this information.

A number is chosen at random from the universal set $\mathcal{E}$
(b) Find the probability that the number is in the set $A \cap B$

$A \cap B = \{6, 9\} \\[3ex] A \cup B = \{1, 2, 5, 6, 8, 9\} \\[3ex] (A \cup B)' = \{3, 4, 7\} \\[3ex]$ (a) The Venn diagram is:

$(b) \\[3ex] n(A \cap B) = 2 \\[3ex] n(\mathcal{E}) = 9 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(\mathcal{E})} = \dfrac{2}{9}$
(89.) WASCCE In a class of 40 students, 18 passed Mathematics, 19 passed Accounts, 16 passed Economics, 5 passed Mathematics and Accounts only, 6 Mathematics only, 9 Accounts only, 2 Accounts and Economics only.
If each student offered at least one of the subjects,

(a) how many students failed in all subjects?

(b) find the percentage number that failed in at least one of Economics and Mathematics

(c) calculate the probability that a student picked at random failed in Accounts.

$n(\mu) = 40 \\[3ex] Let: \\[3ex] Mathematics = M \\[3ex] Accounts = A \\[3ex] Economics = E \\[3ex] n(M) = 18 \\[3ex] n(A) = 19 \\[3ex] n(E) = 16 \\[3ex] n(M\;\;and\;\;A\;\;only) = 5 \\[3ex] n(M\;\;only) = 6 \\[3ex] n(A\;\;only) = 9 \\[3ex] n(A\;\;and\;\;E\;\;only) = 2 \\[3ex] n(M\;\;and\;\;A\;\;and\;\;E) = n(passed\;\;all\;\;three) \\[3ex]$ Let us use a Venn diagram

$\underline{Accounts} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;A\;\;only) + n(A\;\;and\;\;E\;\;only) + n(A\;\;only) = n(A) \\[3ex] n(passed\;\;all\;\;three) + 5 + 2 + 9 = 19 \\[3ex] n(passed\;\;all\;\;three) + 16 = 19 \\[3ex] n(passed\;\;all\;\;three) = 19 - 16 \\[3ex] n(passed\;\;all\;\;three) = 3 \\[3ex] \underline{Mathematics} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;A\;\;only) + n(M\;\;and\;\;E\;\;only) + n(M\;\;only) = n(M) \\[3ex] 3 + 5 + n(M\;\;and\;\;E\;\;only) + 6 = 18 \\[3ex] 14 + n(M\;\;and\;\;E\;\;only) = 18 n(M\;\;and\;\;E\;\;only) = 18 - 14 \\[3ex] n(M\;\;and\;\;E\;\;only) = 4 \\[3ex] \underline{Economics} \\[3ex] n(passed\;\;all\;\;three) + n(M\;\;and\;\;E\;\;only) + n(A\;\;and\;\;E\;\;only) + n(E\;\;only) = n(E) \\[3ex] 3 + 4 + 2 + n(E\;\;only) = 16 \\[3ex] 9 + n(E\;\;only) = 16 \\[3ex] n(E\;\;only) = 16 - 9 \\[3ex] n(E\;\;only) = 7 \\[3ex] \implies \\[3ex] (a) \\[3ex] n(M\;\;only) + n(A\;\;only) + n(E\;\;only) \\[3ex] + n(M\;\;and\;\;A\;\;only) + n(N\;\;and\;\;E\;\;only) + n(A\;\;and\;\;E\;\;only) \\[3ex] + n(passed\;\;all\;\;three) + n(failed\;\;all\;\;three) = n(\mu) \\[3ex] 6 + 9 + 7 + 5 + 4 + 2 + 3 + n(failed\;\;all\;\;three) = 40 \\[3ex] 36 + n(failed\;\;all\;\;three) = 40 \\[3ex] n(failed\;\;all\;\;three) = 40 - 36 \\[3ex] n(failed\;\;all\;\;three) = 4 \\[3ex]$ 4 students failed all subjects.

The completed Venn diagram is:

(b) Failed at least one of Economics and Mathematics implies:
failed in Economics only
failed in Mathematics only
failed in Economics and Mathematics only

failed in all three subjects

In other words, it implies: not passing anything that has to do with Economics or Mathematics
Because of the Venn diagram (that shows passing for the subjects):
This means:
passing only Accounts
failing all three subjects

$n(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics) \\[3ex] = n(A\;\;only) + n(failed\;\;all\;\;three) \\[3ex] = 9 + 4 \\[3ex] = 13 \\[3ex] n(U) = 40 \\[3ex] \%(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics) \\[3ex] = \dfrac{n(failed\;\;in\;\;at\;\;least\;\;one\;\;of\;\;Economics\;\;and\;\;Mathematics)}{n(U)} * 100 \\[5ex] = \dfrac{13}{40} * 100 \\[5ex] = 32.5\% \\[3ex] (c) \\[3ex] P(failed\;\;in\;\;Accounts) + P(passed\;\;in\;\;Accounts) = 1 ...Complementary\;\;Rule \\[3ex] P(failed\;\;in\;\;Accounts) = 1 - P(passed\;\;in\;\;Accounts) \\[3ex] P(passed\;\;in\;\;Accounts) = \dfrac{n(A)}{n(U)} = \dfrac{19}{40} \\[5ex] \implies \\[3ex] P(passed\;\;in\;\;Accounts) = 1 - \dfrac{19}{40} = \dfrac{40 - 19}{40} = \dfrac{21}{40} \\[5ex]$ Alternatively:
Failed in Accounts means that the student did not pass anything related to Accounts
It is $n(A')$
Because of the Venn diagram (that shows passing for the subjects):
This includes:
passing only Mathematics
passing only Economics
passed Mathematics and Economics only
failing all three subjects

$n(failed\;\;in\;\;Accounts) \\[3ex] = n(M\;\;only) + n(E\;\;only) + n(M\;\;and\;\;E\;\;only) + n(failed\;\;all\;\;three) \\[3ex] = 6 + 7 + 4 + 4 \\[3ex] = 21 \\[3ex] n(U) = 40 \\[3ex] P(failed\;\;in\;\;Accounts) \\[3ex] = \dfrac{n(failed\;\;in\;\;Accounts)}{n(U)} \\[5ex] = \dfrac{21}{40}$
(90.) ACT About $1.48 * 10^8$ square kilometers of Earth's surface is land; the rest, about $3.63 * 10^8$ square kilometers, is water.
If a returning space capsule lands at a random point on Earth's surface, which of the following is the best estimate of the probability that the space capsule will land in water?

$F.\;\; 80\% \\[3ex] G.\;\; 71\% \\[3ex] H.\;\; 65\% \\[3ex] J.\;\; 41\% \\[3ex] K.\;\; 29\% \\[3ex]$

$Let: \\[3ex] Land = L \\[3ex] Water = W \\[3ex] n(L) = 1.48 * 10^8 \\[3ex] n(W) = 3.63 * 10^8 \\[3ex] n(S) = n(L) + n(W) \\[3ex] = 1.48 * 10^8 + 3.63 * 10^8 \\[3ex] = 10^8(1.48 + 3.63) \\[3ex] = 10^8 * 5.11 \\[3ex] = 5.11 * 10^8 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} \\[5ex] = \dfrac{3.63 * 10^8}{5.11 * 10^8} \\[5ex] = 0.71037182 \\[3ex] = 0.71037182 * 100 \\[3ex] = 71.037182\% \\[3ex] \approx 71\%$
(91.) ACT A basket contains 10 solid-colored balls - 2 blue, 3 red, and 5 green.
Each ball has a single number printed on it.
The blue balls are numbered 1 and 2 (each number is used once), the red balls are numbered 1 - 3 (each number is used once), and the green balls are numbered 1 - 5 (each number is used once).
A ball will be drawn at random from the basket.
What is the probability that the ball that is drawn will be red or have a 3 printed on it?

$F.\;\; \dfrac{1}{10} \\[5ex] G.\;\; \dfrac{2}{10} \\[5ex] H.\;\; \dfrac{3}{10} \\[5ex] J.\;\; \dfrac{4}{10} \\[5ex] K.\;\; \dfrac{5}{10} \\[5ex]$

$Let\:\: Blue = B \\[3ex] Red = R \\[3ex] Green = G \\[3ex] Numbers\:\: as\:\: is \\[3ex] S = \{1B, 2B, 1R, 2R, 3R, 1G, 2G, 3G, 4G, 5G\} \\[3ex] n(S) = 10 \\[3ex] n(R) = 3 \:\: \implies (1R, 2R, 3R) \\[3ex] n(3) = 2 \:\: \implies (3R, 3G) \\[3ex] n(3 \:\:AND\:\: R) = n(3R) = 1 \\[3ex] P(3 \:\:OR\:\: R) = P(R) + P(3) - P(3 \:\:AND\:\: R)...Addition\:\: Rule \\[3ex] P(3 \:\:OR\:\: R) = \dfrac{3}{10} + \dfrac{2}{10} - \dfrac{1}{10} \\[5ex] = \dfrac{3 + 2 - 1}{10} \\[5ex] = \dfrac{4}{10}...Option\;J \\[5ex] = \dfrac{2}{5}$
(92.) WASSCE A die was rolled a number of times.
The outcomes are as shown in the table.

 Number $1$ $2$ $3$ $4$ $5$ $6$ Outcomes $32$ $m$ $25$ $40$ $28$ $45$

If the probability of obtaining 2 is 0.15, find the:
(a.) value of m
(b.) number of times the die was rolled
(c.) probability of obtaining an even number

$n(S) = n(Outcomes) \\[3ex] = 32 + m + 25 + 40 + 28 + 45 \\[3ex] = m + 170 \\[3ex] (a.) \\[3ex] n(2) = m \\[3ex] P(2) = \dfrac{n(2)}{n(S)} \\[5ex] 0.15 = \dfrac{m}{m + 170} \\[5ex] 0.15(m + 170) = m \\[3ex] 0.15m + 25.5 = m \\[3ex] m = 0.15m + 25.5 \\[3ex] m - 0.15m = 25.5 \\[3ex] 0.85m = 25.5 \\[3ex] m = \dfrac{25.5}{0.85} \\[5ex] m = 30 \\[3ex] (b.) \\[3ex] Number\;\;of\;\;times\;\;the\;\;die\;\;was\;\;rolled \\[3ex] = m + 170 \\[3ex] = 30 + 170 \\[3ex] = 200\;times \\[3ex] (c.) \\[3ex] P(even\;\;number) \\[3ex] = P(2)\;\;OR\;\;P(4)\;\;OR\;\;P(6) \\[3ex] = 0.15 + \dfrac{n(4)}{n(S)} + \dfrac{n(6)}{n(S)} \\[5ex] = 0.15 + \dfrac{40}{200} + \dfrac{45}{200} \\[5ex] = 0.15 + 0.2 + 0.225 \\[3ex] = 0.575$
(93.)

(94.) USSCE - Advance Mathematics Paper 1 Two events are mutually exclusive if on one trial of an experiment:
A. both must occur
B. exactly one must occur
C. exactly one may occur
D. both may occur

When two events are mutually exclusive, both events cannot occur at the same time.
Exactly one event must occur

This means that the first event must occur or the second event must occur.
Both events cannot occur simultaneously.

If we say that exactly one event may occur, then there is the possibility that none of the events occur.
This goes against the formula for mutually exclusive events
Assume events A and B are two mutually exclusive events

$P(A \;\;OR\;\; B) = P(A) + P(B) - P(A \;\;AND\;\; B) \\[3ex] P(A \cup B) = P(A) + P(B) - P(A \cap B) \\[3ex] But\;\; P(A \cap B) = 0...mutually\;\;exclusive\;\;events \\[3ex] \implies \\[3ex] P(A \cup B) = P(A) + P(B) \\[3ex]$ The formula does not take into account that none of the events may occur
Hence, only one event must occur.
(95.) WASCCE In a class of 200 students, 70 offered Physics, 90 Chemistry, 100 Mathematics while 24 did not offer any of the three subjects.
Twenty three (23) students offered Physics and Chemistry, 41 Chemistry and Mathematics while 8 offered all three subjects.

(a) Draw a Venn diagram to illustrate the information.

(b) Find the probability that a student selected at random from the class offered:
(i) Physics only
(ii) Exactly two of the subjects

$n(\mu) = 200 \\[3ex] Let: \\[3ex] Physics = P \\[3ex] Chemistry = C \\[3ex] Mathematics = M \\[3ex] n(P) = 70 \\[3ex] n(C) = 90 \\[3ex] n(M) = 100 \\[3ex] n(neither) = 24 \\[3ex] n(P\;\;and\;\;C) = 23 \\[3ex] n(C\;\;and\;\;M) = 41 \\[3ex] n(all\;\;three) = 8 \\[3ex] n(P\;\;and\;\;M\;\;only) = d \\[3ex]$ (a) The Venn diagram is:

$n(P\;\;and\;\;C\;\;only) \\[3ex] = 23 - 8 \\[3ex] = 15 \\[3ex] n(C\;\;and\;\;M\;\;only) \\[3ex] = 41 - 8 \\[3ex] = 33 \\[3ex] n(C\;\;only) \\[3ex] = 90 - (15 + 8 + 33) \\[3ex] = 90 - 56 \\[3ex] = 34 \\[3ex] n(P\;\;only) \\[3ex] = 70 - (d + 8 + 15) \\[3ex] = 70 - (d + 23) \\[3ex] = 70 - d - 23 \\[3ex] = 47 - d \\[3ex] n(M\;\;only) \\[3ex] = 100 - (d + 8 + 33) \\[3ex] = 100 - (d + 41) \\[3ex] = 100 - d - 41 \\[3ex] = 59 - d \\[3ex] \implies \\[3ex] (47 - d) + (59 - d) + 34 + d + 15 + 33 + 8 + 24 = 200 \\[3ex] 47 - d + 59 - d + d + 34 + 15 + 33 + 8 + 24 = 200 \\[3ex] 220 - d = 200 \\[3ex] 220 - 200 = d \\[3ex] 20 = d \\[3ex] d = 20 \\[3ex] \implies \\[3ex] n(P\;\;and\;\;M\;\;only) = 20 \\[3ex] (b) \\[3ex] (i) \\[3ex] n(P\;\;only) = 47 - d = 47 - 20 = 27 \\[3ex] n(\mu) = 200 \\[3ex] P(P\;\;only) = \dfrac{n(P\;\;only)}{n(\mu)} = \dfrac{27}{200} \\[5ex] (ii) \\[3ex] n(exactly\;\;2\;\;subjects) \\[3ex] = n(P\;\;and\;\;C\;\;only) + n(C\;\;and\;\;M\;\;only) + n(P\;\;and\;\;M\;\;only) \\[3ex] = 15 + 33 + 20 \\[3ex] = 68 \\[3ex] P(exactly\;\;2\;\;subjects) \\[3ex] = \dfrac{n(exactly\;\;2\;\;subjects)}{n(\mu)} \\[5ex] = \dfrac{68}{200} \\[5ex] = \dfrac{17}{50}$
(96.) GCSE $\mathcal{E}$ = {even numbers between 1 and 25}
A = {2, 8, 10, 14}
B = {6, 8, 20}
C = {8, 18, 20, 22}

(a) Complete the Venn diagram for this information.

A number is chosen at random from $\mathcal{E}$
(b) Find the probability that the number is a member of $A \cap B$

$\mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} \\[3ex] A = \{2, 8, 10, 14\} \\[3ex] B = \{6, 8, 20\} \\[3ex] C = \{8, 18, 20, 22\} \\[3ex] n(all\;\;three) = n(A \cap B \cap C) = \{8\} \\[3ex] A' = \{4, 6, 12, 16, 18, 20, 22, 24\} \\[3ex] B' = \{2, 4, 10, 12, 14, 16, 18, 22, 24\} \\[3ex] C' = \{2, 4, 6, 10, 12, 14, 16, 24\} \\[3ex] \underline{A\;\;and\;\;B\;\;only} \\[3ex] A \cap B = \{8\} \\[3ex] A \cap B \cap C' = \{\} \\[3ex] \underline{A\;\;and\;\;C\;\;only} \\[3ex] A \cap C = \{8\} \\[3ex] A \cap C \cap B' = \phi \\[3ex] \underline{B\;\;and\;\;C\;\;only} \\[3ex] B \cap C = \{8, 20\} \\[3ex] B \cap C \cap A' = \{20\} \\[3ex] \underline{A\;\;only} \\[3ex] B' \cap C' = \{2, 4, 10, 12, 14, 16, 24\} \\[3ex] A \cap B' \cap C' = \{2, 10, 14\} \\[3ex] \underline{B\;\;only} \\[3ex] A' \cap C' = \{4, 6, 12, 16, 24\} \\[3ex] B \cap A' \cap C' = \{6\} \\[3ex] \underline{C\;\;only} \\[3ex] A' \cap B' = \{4, 12, 16, 18, 22, 24\} \\[3ex] C \cap A' \cap B' = \{18, 22\} \\[3ex]$ (a) The Venn diagram is:

$A \cap B = \{8\} \\[3ex] n(A \cap B) = 1 \\[3ex] \mathcal{E} = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24\} \\[3ex] n(\mathcal{E}) = 12 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(\mathcal{E})} = \dfrac{1}{12}$
(97.)

(98.) WASSCE If two numbers are selected at random, one after the other, with replacement from the set A = {5, 6, 7, 8, 9}, find the probability of selecting at least one prime number.

Two numbers are selected: With Replacement condition
Prime numbers = {5, 7}
At least one prime number means one or two prime numbers for example: {5, 5}, {5, 6}, {5, 7}, etc.
Because there are several options involved, it is better to use a Punnett Square
The set containing at least one prime number is marked in red.

First Number in the Column and Second Number in the Row
$2nd\:\:Number\:\:\rightarrow$
$1st\:\:Number\:\:\downarrow$
$5$ $6$ $7$ $8$ $9$
$5$ $\color{red}{5, 5}$ $\color{red}{5, 6}$ $\color{red}{5, 7}$ $\color{red}{5, 8}$ $\color{red}{5, 9}$
$6$ $\color{red}{6, 5}$ $6, 6$ $\color{red}{6, 7}$ $6, 8$ $6, 9$
$7$ $\color{red}{7, 5}$ $\color{red}{7, 6}$ $\color{red}{7, 7}$ $\color{red}{7, 8}$ $\color{red}{7, 9}$
$8$ $\color{red}{8, 5}$ $8, 6$ $\color{red}{8, 7}$ $8, 8$ $8, 9$
$9$ $\color{red}{9, 5}$ $9, 6$ $\color{red}{9, 7}$ $9, 8$ $9, 9$

$n(at\;\;least\;\;one\;\;prime\;\;number) = n(E) = 16 \\[3ex] n(S) = 5 * 5 = 25 \\[3ex] P(at\;\;least\;\;one\;\;prime\;\;number) \\[3ex] = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{16}{25}$
(99.) HSC Mathematics Standard 1 Barbara plays a game of chance, in which two unbiased six-sided dice are rolled.
The score for the game is obtained by finding the difference between the two numbers rolled.
For example, if Barbara rolls a 2 and a 5, the score is 3.
The table shows some of the scores.

(a.) Complete the six missing values in the table to show all possible scores for the game.
(b.) What is the probability that the score for a game is NOT 0?

(a.)
The missing values are marked in red
Absolute Value of:
$Die\:\:1\;\rightarrow \\[1ex] \;\;\;- \\[1ex] Die\:\:2\;\downarrow$ $1$ $2$ $3$ $4$ $5$ $6$
$1$ $0$ $1$ $2$ $3$ $4$ $5$
$2$ $\color{red}1$ $0$ $1$ $2$ $3$ $4$
$3$ $2$ $1$ $\color{red}0$ $\color{red}1$ $2$ $\color{red}3$
$4$ $3$ $2$ $1$ $\color{red}0$ $1$ $2$
$5$ $\color{red}4$ $3$ $2$ $1$ $0$ $1$
$6$ $5$ $4$ $3$ $2$ $1$ $0$

$(b.) \\[3ex] P(NOT\;\;0) + P(0) = 1...Complementary\;\;Rule \\[3ex] P(0) = \dfrac{n(0)}{n(S)} \\[5ex] n(S) = 6 * 6 = 36 \\[3ex] n(0) = 6 \\[3ex] P(0) = \dfrac{6}{36} = \dfrac{1}{6} \\[5ex] P(NOT\;\;0) \\[3ex] = 1 - P(0) \\[5ex] = 1 - \dfrac{1}{6} \\[5ex] = \dfrac{6}{6} - \dfrac{1}{6} \\[5ex] = \dfrac{5}{6}$
(100.) ACT A bag contains 8 red marbles, 9 yellow marbles, and 7 green marbles.
How many additional red marbles must be added to the 24 marbles already in the bag so that the probability of randomly drawing a red marble is $\dfrac{3}{5}$?

$A.\;\; 11 \\[3ex] B.\;\; 16 \\[3ex] C.\;\; 20 \\[3ex] D.\;\; 24 \\[3ex] E.\;\; 32 \\[3ex]$ Before you solve this question, check out Question (8.)
Did you notice anything?

We can solve this question in two ways
Use any method you prefer

Let the number of additional red marbles = $r$

$\underline{First\:\:Approach} \\[3ex] \underline{Initial} \\[3ex] n(S) = 24 \\[3ex] n(Red) = 8 \\[3ex] \underline{Updated} \\[3ex] n(Red) = 8 + r \\[3ex] n(S) = 24 + r \\[3ex] P(Red) = \dfrac{8 + r}{24 + r} \\[5ex] P(Red) = \dfrac{3}{5} \\[5ex] \therefore \dfrac{8 + r}{24 + r} = \dfrac{3}{5} \\[5ex] Cross\:\: Multiply\:\: method \\[3ex] 5(8 + r) = 3(24 + r) \\[3ex] 40 + 5r = 72 + 3r \\[3ex] 5r - 3r = 72 - 40 \\[3ex] 2r = 32 \\[3ex] r = \dfrac{32}{2} \\[5ex] r = 16 \\[5ex] \underline{Second\:\:Approach} \\[3ex] P(Red) = \dfrac{3}{5} \\[5ex] P(Red) + P(Red') = 1 ...Complementary\:\:Rule \\[3ex] P(Red') = 1 - P(Red) \\[3ex] P(Red') = 1 - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5}{5} - \dfrac{3}{5} \\[5ex] P(Red') = \dfrac{5 - 3}{5} \\[5ex] P(Red') = \dfrac{2}{5} \\[5ex] n(Red') = n(Yellow) + n(Green) \\[3ex] n(Red') = 9 + 7 = 16 \\[3ex] P(Red') = \dfrac{n(Red')}{n(S)} \\[5ex] n(S) * P(Red') = n(Red') \\[3ex] n(S) = \dfrac{n(Red')}{P(Red')} \\[5ex] n(S) = n(Red') \div P(Red') \\[3ex] n(S) = 16 \div \dfrac{2}{5} \\[5ex] n(S) = 16 * \dfrac{5}{2} \\[5ex] n(S) = 8(5) \\[3ex] n(S) = 40 \\[3ex] Additional\:\:red\:\:balls = 40 - (8 + 9 + 7) \\[3ex] Additional\:\:red\:\:balls = 40 - 24 \\[3ex] Additional\:\:red\:\:balls = 16 \\[3ex]$ $16$ red marbles need to be added so that the probability of drawing a red marble is $\dfrac{3}{5}$

(101.)

(102.) WASSCE A fair die is thrown two times.
What is the probability that the sum of the scores is at least 10?

$A.\;\; \dfrac{5}{36} \\[5ex] B.\;\; \dfrac{1}{6} \\[5ex] C.\;\; \dfrac{5}{18} \\[5ex] D.\;\; \dfrac{2}{3} \\[5ex]$

Sample Space for Two Throws of a Fair Die
First Throw in the Column and Second Throw in the Row
$2nd\:\:Throw\:\:\rightarrow$
($+$)
$1st\:\:Throw\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $2$ $3$ $4$ $5$ $6$ $7$
$2$ $3$ $4$ $5$ $6$ $7$ $8$
$3$ $4$ $5$ $6$ $7$ $8$ $9$
$4$ $5$ $6$ $7$ $8$ $9$ $\color{red}{10}$
$5$ $6$ $7$ $8$ $9$ $\color{red}{10}$ $\color{red}{11}$
$6$ $7$ $8$ $9$ $\color{red}{10}$ $\color{red}{11}$ $\color{red}{12}$

$n(S) = 6 * 6 = 36 \\[3ex] Let\;\;E = event\;\;of\;\;selecting\;\;at\;\;least\;\;10 \\[3ex] E = \{10, 10, 11, 10, 11, 12\} \\[3ex] n(E) = 6 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{6}{36} \\[5ex] = \dfrac{1}{6}$
(103.)

(104.) GCSE The Venn diagram shows information about 50 people who are in bands.

(a) How many of the people are guitar players?

(b) How many of the people are singers but not guitar players?

(c) One of the people is chosen at random.
Write down the probability that the person is not a singer and not a guitar player.

$(a) \\[3ex] n(guitar\;\;players) = n(G) = 21 + 8 = 29 \\[3ex]$ 29 people are guitar players

$(b) \\[3ex] n(singers\;\;but\;\;not\;\;guitar\;\;players) \\[3ex] = n(singers\;\;only) \\[3ex] = 4 \\[3ex]$ 4 people are singers but not guitar players

$(c) \\[3ex] n(universal\;\;set) = n(\xi) = 50 \\[3ex] n(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player) = 17 \\[3ex] P(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player) \\[3ex] = \dfrac{n(not\;\;a\;\;singer\;\;and\;\;not\;\;a\;\;guitar\;\;player)}{n(\xi)} \\[5ex] = \dfrac{17}{50}$

ACT Use the following information to answer Questions 105 - 108.
Many humans carry the gene Yq77.
The Yq test determines, with 100% accuracy, whether a human carries Yq77.
If a Yq test result is negative, the human does NOT carry Yq77
Sam designed a less expensive test for Yq77 called the Sam77 test.
It produces some incorrect results.
To determine the accuracy of the Sam77 test, both tests were administered to 1,000 volunteers.
The results from this administration are summarized in the table below.

 Positive Yq test Negative Yq test Positive Sam77 test Negative Sam77 test 590 25 10 375

(105.) It cost $2,500 to administer each Yq test and$50 to administer each Sam77 test.
What was the total cost to administer both tests to all the volunteers?

$A.\;\; \$1,537,500 \\[3ex] B.\;\; \$1,556,750 \\[3ex] C.\;\; \$1,568,250 \\[3ex] D.\;\; \$2,500,000 \\[3ex] E.\;\; \$2,550,000 \\[3ex]  Number\;\;of\;\;participants = 1000 \\[3ex] 1000\;\;Yq77\;\;tests\;\;@\;\;\$2500/test = 1000(2500) = 2,500,000 \\[3ex] 1000\;\;Sam77\;\;tests\;\;@\;\;\$50/test = 1000(50) = 50,000 \\[3ex] Total\;\;cost = \$2,500,000 + \$50,000 = \$2,550,000$
(106.) What percent of the volunteers actually carry Yq77?

$F.\;\; 57.5\% \\[3ex] G.\;\; 60.0\% \\[3ex] H.\;\; 60.5\% \\[3ex] J.\;\; 61.5\% \\[3ex] K.\;\; 62.5\% \\[3ex]$

$Number\;\;of\;\;participants = 1000 \\[3ex] Number\;\;of\;\;positive\;\;Yq77\;\;test = 590 + 25 = 615 \\[3ex] \%\;\;of\;\;positive\;\;Yq77\;\;test \\[3ex] = \dfrac{615}{1000} * 100 \\[5ex] = 61.5\%$
(107.) For how many volunteers did the Sam77 test give an incorrect result?

$A.\;\; 10 \\[3ex] B.\;\; 25 \\[3ex] C.\;\; 35 \\[3ex] D.\;\; 385 \\[3ex] E.\;\; 400 \\[3ex]$

Yq77 test is 100% accurate but Sam77 test is not.
So, an incorrect result by the Sam77 test implies that the:
(a.) volunteer had a positive Sam77 test but a negative Yq77 test (10 volunteers)
(b.) volunteer had a negative Sam77 test but a positive Yq77 test (25 volunteers)

10 + 25 = 35
This means that 35 volunteers received an incorrect result from the Sam77 test.
(108.) One of the volunteers whose Sam77 test result was positive will be chosen at random.
To the nearest 0.001, what is the probability the chosen volunteer does NOT possess Yq77?

$F.\;\; 0.017 \\[3ex] G.\;\; 0.026 \\[3ex] H.\;\; 0.035 \\[3ex] J.\;\; 0.041 \\[3ex] K.\;\; 0.063 \\[3ex]$

The question deals with Conditional Probability.
It can be reworded as:
What is the probability that the chosen volunteer does not possess Yq77 given that the Sam77 test result is positive?

$Let: \\[3ex] positive\;\;Sam77\;\;test = A \\[3ex] positive\;\;Yq77\;\;test = Y \\[3ex] \implies negative\;\;Yq77\;\;test = Y' \\[3ex] P(Y' | A) = \dfrac{n(Y' \cap A)}{n(A)} \\[5ex] = \dfrac{10}{590 + 10} \\[5ex] = \dfrac{10}{600} \\[5ex] = 0.0166666667 \\[3ex] \approx 0.017$
(109.) HSC Mathematics Standard 1 In a bag, there are six playing cards, 2, 4, 6, 8, Queen and King.
The Queen and King are known as picture cards.
Two of these cards are chosen randomly.
All of the possible outcomes are shown.

(a.) What is the probability that the two cards chosen include one or both picture cards?
(b.) What is the probability that the two cards chosen do NOT include any picture cards?

(a.)
One or both picture cards mean at least one picture card
Let event E = At least one picture card are:
2 and Queen, 2 and King, 4 and Queen, 4 and King, 6 and Queen, 6 and King, 8 and Queen, 8 and King, Queen and King

$n(S) = 15 \\[3ex] n(E) = 9 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{9}{15} \\[5ex] = \dfrac{3}{5} \\[5ex]$ (b.)
We can solve this question using at least thwo approaches
Use any approach you prefer.

First Approach
Event E = selecting at least one picture card
Event E' = selecting no picture card
Both are complementary events.

$P(E') + P(E) = 1...Complementary\;\;Rule \\[3ex] P(E') = 1 - P(E) \\[3ex] = 1 - \dfrac{3}{5} \\[5ex] = \dfrac{5}{5} - \dfrac{3}{5} \\[5ex] = \dfrac{5 - 3}{5} \\[5ex] = \dfrac{2}{5} \\[5ex]$ Second Approach
Event E' = selecting no picture card

$n(E') = 6 \\[3ex ] P(E') = \dfrac{n(E')}{n(S)} \\[5ex] = \dfrac{6}{15} \\[5ex] = \dfrac{2}{5}$
(110.) WASSCE A bag contains 5 red and 4 blue identical balls.
If 2 balls are selected at random from the bag, one after the other, with replacement, find the probability that the first is red and the second blue.

$A.\;\; \dfrac{2}{9} \\[5ex] B.\;\; \dfrac{5}{18} \\[5ex] C.\;\; \dfrac{20}{81} \\[5ex] D.\;\; \dfrac{5}{9} \\[5ex]$

$Let: \\[3ex] red = R \\[3ex] blue = B \\[3ex] n(R) = 5 \\[3ex] n(B) = 4 \\[3ex] n(S) = 5 + 4 = 9 \\[3ex] \underline{With\;\;Replacement\;\;condition} \\[3ex] P(R\;\;AND\;\;B) \\[3ex] = P(R) * P(B) \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{n(R)}{n(S)} * \dfrac{n(B)}{n(S)} \\[5ex] = \dfrac{5}{9} * \dfrac{4}{9} \\[5ex] = \dfrac{20}{81}$
(111.)

(112.) Identify whether these probabilities are theoretical or empirical.

(a.) A Monopoly player claims that the probability of getting a 3 when rolling a six-sided die is $\dfrac{1}{6}$ because the die is equally likely to land on any of the six sides.

(b.) A person was trying to figure out the probability of getting two heads when flipping two coins.
She flipped two coins 20 times, and in 6 of these 20 times, both coins landed heads.
On the basis of this outcome, she claims that the probability of two heads is $\dfrac{6}{20}$, or 30%.

(c.) A friend flips a coin 10 times and says that the probability of getting a head is 60% because he got six heads.

(d.) A magician claims that he has a fair coin: "fair" because both sides, heads and tails, are equally likely to land face up when the coin is flipped.
He says that if the coin is flipped three times, the probability of getting three heads is $\dfrac{1}{8}$

(a.) This is an example of a theoretical probability because the event is not based on an experiment.

(b.) This is an example of an empirical probability because it is based on an experiment.

(c.) This is an example of an empirical probability because it is based on an experiment.

(d.) This is a theoretical probability. It is not based on an experiment.

$Let\;\;event\;\;of\;\;head\;\;facing\;\;up = H \\[3ex] P(H) = \dfrac{1}{2} \\[5ex] P(HHH) = \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2} \\[5ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;events \\[3ex] = \dfrac{1}{8}$
(113.) ACT The table indicates the grade (10 or 11) and high school (North or South) of the 270 students enrolled in Algebra II in the Green City School District.

Suppose 2 of these students will be chosen at arndom to represent the Algebra II classes at a local STEM (Science, Technology, Engineering, and Mathematics) event.
Which of the following expressions gives the probability that both chosen students will be from the same grade and the same high school?

$A.\;\; \dfrac{47(46)}{270(269)} + \dfrac{93(92)}{270(269)} + \dfrac{73(72)}{270(269)} + \dfrac{57(56)}{270(269)} \\[5ex] B.\;\; \dfrac{1}{4}\left(\dfrac{47}{270} + \dfrac{93}{270} + \dfrac{73}{270} + \dfrac{56}{270}\right) \\[5ex] C.\;\; \dfrac{47(73)}{270(269)} + \dfrac{93(57)}{270(269)} \\[5ex] D.\;\; \dfrac{47(93)}{270(269)} + \dfrac{73(57)}{270(269)} \\[5ex] E.\;\; \dfrac{1}{4}\left(\dfrac{1}{4}\right) \\[5ex]$

(a.) Choosing both students from the same grade and the same high school implies selecting:
Two 10th-grade students from the North high school
OR
Two 10th-grade students from the South high school
OR
Two 11th-grade students from the North high school
OR
Two 11th-grade students from the South high school

(b.) Because both students are chosen simultaneously at random, this is a case of without replacement condition

$n(S) = 47 + 93 + 73 + 57 = 270 \\[3ex] P(choosing\;\;2\;\;students\;\;from\;\;same\;\;grade\;\;and\;\;same\;\;high\;\;school) \\[3ex] = \dfrac{47(46)}{270(269)} + \dfrac{93(92)}{270(269)} + \dfrac{73(72)}{270(269)} + \dfrac{57(56)}{270(269)}$
(114.) NSC
(114.1) Two events, A and B, are such that:
• Events A and B are independent
• P(not A) = 0,4
• P(B) = 0,3
Calculate P(A and B)

(114.2) A survey was conducted among 150 learners at a school.
• The probability that a learner, selected at random, will take part in:
• Only hockey (H) is 0,24
• Hockey and debating (D), but not chess (C) is 0,14
• Debating and chess, but not hockey is 0,12
• Hockey and chess, but not debating is 0,02
• The probability that a learner, selected at random, participates in at least one activity is 0,7
• 15 learners participated in all three activities.
• The number of learners that participate only in debating is the same as the number of learners who participate only in chess.

The Venn diagram below shows some of the above information.

(114.2.1) Determine a, the probability that a learner, selected at random, participates in all three activities.

(114.2.2) Determine m, the probability that a learner, selected at random, does NOT participate in any of the three activities.

(114.2.3) How many learners play only chess?

$(114.1.1) \\[3ex] P(A) + P(not\;\;A) = 1 ...Complementary\;\;Rule \\[3ex] P(A) + 0.4 = 1 \\[3ex] P(A) = 1 - 0.4 \\[3ex] P(A) = 0.6 \\[3ex] P(A\;\;and\;\;B) = P(A) * P(B) \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = 0.6 * 0.3 \\[3ex] = 0.18 \\[5ex] (114.2.1) \\[3ex] a = \dfrac{n(H \cap D \cap C)}{n(S)} \\[5ex] = \dfrac{15}{150} = \dfrac{1}{10} = 0.1 \\[3ex] (114.2.2) \\[3ex] P(at\;\;least\;\;one\;\;activity) + P(no\;\;activity) = 1 \\[3ex] ...Complementary\;\;Rule \\[3ex] 0.7 + m = 1 \\[3ex] m = 1 - 0.7 \\[3ex] m = 0.3 \\[3ex] (114.2.3) \\[3ex] P(only\;\;1\;\;activity) = 0.24 + b + b = 2b + 0.24 \\[3ex] P(only\;\;2\;\;activities) = 0.02 + 0.14 + 0.12 = 0.28 \\[3ex] P(all\;\;three\;\;activities) = a = 0.1 \\[3ex] P(no\;\;activity) = m = 0.3 \\[3ex] P(total\;\;probability) = 1 \\[3ex] \implies \\[3ex] 2b + 0.24 + 0.28 + 0.1 + 0.3 = 1 \\[3ex] 2b + 0.92 = 1 \\[3ex] 2b = 1 - 0.92 \\[3ex] 2b = 0.08 \\[3ex] b = \dfrac{0.08}{2} \\[5ex] b = 0.04 \\[3ex] P(only\;\;chess) = \dfrac{n(only\;\;chess)}{n(S)} \\[5ex] b = \dfrac{n(only\;\;chess)}{150} \\[5ex] n(only\;\;chess) \\[3ex] = 150 * b \\[3ex] = 150 * 0.04 \\[3ex] = 6 \\[3ex]$ 6 learners play only chess
(115.)

(116.) ACT A package of candy contains pieces each of which is 1 of 6 possible colors: brown, red, green, yellow, orange, and blue.
In each package, $\dfrac{1}{3}$ of the pieces are brown and the remaining pieces have an even distribution of the other 5 colors.
What is the probability that a piece drawn randomly from the package is red?

$A.\;\; \dfrac{1}{15} \\[5ex] B.\;\; \dfrac{2}{15} \\[5ex] C.\;\; \dfrac{1}{6} \\[5ex] D.\;\; \dfrac{1}{5} \\[5ex] E.\;\; \dfrac{2}{3} \\[5ex]$

$Let\;\;the\;\;number\;\;of\;\;candy\;\;pieces = n \\[3ex] \implies \\[3ex] n(S) = n \\[3ex] n(brown) = \dfrac{1}{3}n \\[5ex] n(other\;\;colors) = remaining \\[3ex] = 1 - \dfrac{1}{3}n = \dfrac{2}{3}n \\[5ex] n(red) \\[3ex] = \dfrac{2}{3}n \div 5...even\;\;distribution \\[5ex] = \dfrac{2}{3}n * \dfrac{1}{5} \\[5ex] = \dfrac{2n}{15} \\[5ex] P(red) = \dfrac{n(red)}{n(S)} \\[5ex] = n(red) \div n(S) \\[3ex] = \dfrac{2n}{15} \div n \\[5ex] = \dfrac{2n}{15} * \dfrac{1}{n} \\[5ex] = \dfrac{2}{15}$
(117.)

(118.) ACT A bag contains 64 marbles, all solid colored.
Each marble is either red, yellow, or green.
A marble is randomly removed from the bag and then returned to the bag.

The probability that this marble is red is $\dfrac{5}{8}$

The probability that this marble is yellow is $\dfrac{1}{4}$

How many green marbles are in the bag?

A marble is randomly removed from the bag and then returned to the bag.
This is a case of "with replacement" - independent events

$Let\:\: red = R \\[3ex] yellow = Y \\[3ex] green = G \\[3ex] n(S) = 64 \\[3ex] P(R) = \dfrac{5}{8} \\[5ex] P(Y) = \dfrac{1}{4} \\[5ex] P(R) = \dfrac{n(R)}{n(S)} \\[5ex] \rightarrow \dfrac{5}{8} = \dfrac{n(R)}{64} \\[5ex] 8 * 8 = 64 \\[3ex] \therefore n(R) = 5 * 8 = 40 \\[3ex] P(Y) = \dfrac{n(Y)}{n(S)} \\[5ex] \rightarrow \dfrac{1}{4} = \dfrac{n(Y)}{64} \\[5ex] 4 * 16 = 64 \\[3ex] \therefore n(Y) = 1 * 16 = 16 \\[3ex] n(G) + n(R) + n(Y) = 64 \\[3ex] n(G) + 40 + 16 = 64 \\[3ex] n(G) + 56 = 64 \\[3ex] n(G) = 64 - 56 \\[3ex] n(G) = 8 \\[3ex]$ There are $8$ green marbles in the bag.
(119.) ACT A certain company has 120 employees, 85 of whom have business degrees.
Of the employees with business degrees, 75 are certified public accountants (CPAs).
There are 14 employees who are not CPAs and also do not hold a business degree.
One employee of the company will be selected at random to be interviewed for a television news program.
What is the probability that the selected employee will be a CPA?
(Note: A business degree is NOT required to be a CPA.)

$F.\;\; \dfrac{75}{120} \\[5ex] G.\;\; \dfrac{85}{120} \\[5ex] H.\;\; \dfrac{89}{120} \\[5ex] J.\;\; \dfrac{96}{120} \\[5ex] K.\;\; \dfrac{99}{120} \\[5ex]$

Let us analyze this question.
Based on the last sentence, it is possible to have CPAs who do not have business degrees.
So, we have:
120 employees, say E (given)
those with business degrees, say B (85 employees...given); and those with non-business degrees, B' (not given)
Those with business degrees consists of:
CPAs, say C (75 employees...given); and non-CPAs, C' (not given)
CPAs, C (not given); and non-CPAs, C' (14...given)

$n(E) = 120...given \\[3ex] n(E) = n(B) + n(B') \\[3ex] n(B) = 85...given \\[3ex] n(C) = 75 ...given \\[3ex] n(B') = n(E) - n(B) = 120 - 85 = 35...found \\[3ex] n(B') = n(C) + n(C') \\[3ex] n(C') = 14...given \\[3ex] n(C) = n(B') - n(C') = 35 - 14 = 21...found \\[3ex]$ Let us represent this information as a tree diagram...visualize it

$n(C) = 75 + 21 ...given + found \\[3ex] n(C) = 96 \\[3ex] n(S) = n(E) = 120 \\[3ex] P(C) = \dfrac{n(C)}{n(S)} \\[5ex] = \dfrac{96}{120}$
(120.) GCSE Twenty people go on a trip to the seaside.
Of these 20 people
• 13 swim in the sea
• 17 go to the funfair
• 2 do not swim in the sea or go to the funfair
(a) Complete the Venn diagram below to show this information.

One person is chosen at random.

(b) Find the probability that this person swims in the sea and goes to the funfair.

(c) Find the probability that this person either swims in the sea or goes to the funfair, but does not do both.

(d.) One person is chosen at random from those who swim in the sea.
Find the probability that this person does not go to the funfair.

$Let: \\[3ex] swim\;\;in\;\;sea = W \\[3ex] go\;\;to\;\;the\;\;funfair = G \\[3ex] n(swim) = n(W) = 13 \\[3ex] n(go) = n(G) = 17 \\[3ex] n(both) = n(W \cap G) = p n(swim\;\;only) = 13 - p \\[3ex] n(go\;\;only) = 17 - p \\[3ex] n(neither) = n((W \cup G)') = 2 \\[3ex]$ The Venn diagram is:

$13 - p + 17 - p + p + 2 = 20 \\[3ex] 32 - p = 20 \\[3ex] 32 - 20 = p \\[3ex] p = 12 \\[3ex] \implies \\[3ex] n(both) = 12 \\[3ex] n(swim\;\;only) \\[3ex] = 13 - p \\[3ex] = 13 - 12 \\[3ex] = 1 \\[3ex] n(go\;\;only) \\[3ex] = 17 - p \\[3ex] = 17 - 12 \\[3ex] = 5 \\[3ex]$ The completed Venn diagram is:

$(b) \\[3ex] P(both) = \dfrac{n(both)}{n(\varepsilon)} \\[5ex] = \dfrac{12}{20} \\[5ex] = \dfrac{3}{5} \\[5ex] (c) \\[3ex]$ Swimming in the sea or going to the funfair without doing both are:
those who do only one: swim the sea only OR go to the funfair only

$n(only\;\;one) \\[3ex] = 1 + 5 = 6 \\[3ex] P(only\;\;one) \\[3ex] = \dfrac{n(only\;\;one)}{n(\varepsilon)} \\[5ex] = \dfrac{6}{20} \\[5ex] = \dfrac{3}{10} \\[5ex]$ Selecting somone who swims but does not go to the funfair means that the person swims only

$(d) \\[3ex] n(swim) = 13 \\[3ex] n(swim\;\;only) = 1 \\[3ex] P(swim\;\;only) = \dfrac{n(swim\;\;only)}{n(swim)} = \dfrac{1}{13}$

(121.)

(122.)

(123.) GCSE There are 135 passengers on a plane.
3 of the passengers in Business Class are flying for the first time.
In total, there are 15 passengers in Business Class.
$\dfrac{1}{4}$ of the passengers not in Business Class are flying for the first time.

(a.) In the Venn diagram,
$\xi$ = passengers on the plane
B = passengers in Business Class
F = passengers flying for the first time
Complete the Venn diagram.

(b.) One of the passengers is chosen at random.
Write down the probability that the passenger is in Business Class.

$(a.) \\[3ex] n(\xi) = 135 \\[3ex] n(B \cap F) = 3 \\[3ex] n(B) = 15 \\[3ex] \implies \\[3ex] n(B\;\;ONLY) = n(B \cap F') \\[3ex] = 15 - 3 = 12 \\[3ex] n(B') = n(\xi) - n(B) \\[3ex] = 135 - 15 \\[3ex] = 120 \\[3ex] n(F \cap B') = \dfrac{1}{4} * 120 = 30 \\[5ex] n(B' \cap F') \\[3ex] = n(neither\;\;B\;\;nor\;\;F) \\[3ex] = 135 - (12 + 30 + 3) \\[3ex] = 135 - 45 \\[3ex] = 90 \\[3ex]$

$(b.) \\[3ex] n(B) = 15 \\[3ex] n(\xi) = 135 \\[3ex] P(B) = \dfrac{n(B)}{n(\xi)} \\[5ex] = \dfrac{15}{135} \\[5ex] = \dfrac{1}{9}$
(124.) HSC Mathematics Standard 2 There are 8 chocolates in a box.
Three have peppermint centres (P) and five have caramel centres (C)
Kim randomly chooses a chocolate from the box and eats it.
Sam then randomly chooses and eats one of the remaining chocolates.
A partially completed probability tree is shown.

What is the probability that Kim and Sam choose chocolates with different centres?

$A.\;\; \dfrac{15}{64} \\[5ex] B.\;\; \dfrac{15}{56} \\[5ex] C.\;\; \dfrac{15}{32} \\[5ex] D.\;\; \dfrac{15}{28} \\[5ex]$

(a.) Kim chooses a chocolate and eats it.
Sam then chooses another chocolate and eats it.
This is a case of two selections, one at a time, without replacement

(b.) Choosing chocolates with different centres implies that:
Kim chose peppermint (P) AND Sam chose caramel (C)
OR
Kim chose caramel (C) AND Sam chose peppermint (P)

$n(P) = 3 \\[3ex] n(C) = 5 \\[3ex] n(S) = 3 + 5 = 8 \\[3ex] \underline{Without\;\;Replacement\;\;condition} \\[3ex] P(P\;\;AND\;\;C) \;\;OR\;\; P(C\;\;AND\;\;P) \\[3ex] = \left(\dfrac{3}{8} * \dfrac{5}{7}\right) + \left(\dfrac{5}{8} * \dfrac{3}{7}\right) \\[5ex] = \dfrac{15}{56} + \dfrac{15}{56} \\[5ex] = \dfrac{15 + 15}{56} \\[5ex] = \dfrac{30}{56} \\[5ex] = \dfrac{15}{28}$
(125.) WASSCE A survey of 40 students showed that 23 students study Mathematics, 5 study Mathematics and Physics, 8 study Chemistry and Mathematics, 5 study Physics and Chemistry and 3 study all the three subjects.
The number of students who study Physics only is twice the number who study Chemistry only.

(a) Find the number of students who study:
(i) only Physics
(ii) only one subject

(b) What is the probability that a student selected at random studies exactly 2 subjects?

$(a) \\[3ex] Let: \\[3ex] Mathematics = M \\[3ex] Physics = P \\[3ex] Chemistry = C \\[3ex] n(Physics\;\;ONLY) = k \\[3ex] n(Chemistry\;\;ONLY) = 2k \\[3ex]$

$(i) \\[3ex] k + 2k + 13 + 2 + 2 + 5 + 3 = 40 \\[3ex] 3k + 25 = 40 \\[3ex] 3k = 40 - 25 \\[3ex] 3k = 15 \\[3ex] k = \dfrac{15}{3} \\[5ex] k = 5 \\[3ex]$ 5 students study only Physics

$(ii) \\[3ex] n(only\;\;one\;\;subject) \\[3ex] = n(only\;\;Mathematics) + n(only\;\;Physics) + n(only\;\;Chemistry) \\[3ex] = 13 + k + 2k \\[3ex] = 13 + 5 + 2(5) \\[3ex] = 18 + 10 \\[3ex] = 28 \\[3ex]$ 28 students study only one subject

$n(exactly\;\;two\;\;subjects) \\[3ex] = n(only\;\;Mathematics\;\;and\;\;Physics) + n(only\;\;Mathematics\;\;and\;\;Chemistry) + n(only\;\;Physics\;\;and\;\;Chemistry) \\[3ex] = 2 + 5 + 2 \\[3ex] = 9 \\[3ex] n(\xi) = 40 \\[3ex] P(exactly\;\;two\;\;subjects) \\[3ex] = \dfrac{n(exactly\;\;two\;\;subjects)}{n(\xi)} \\[5ex] = \dfrac{9}{40}$
(126.) ACT One day will be randomly selected from the 7 days in a week.
Then 1 month will be randomly selected from the 12 months in a year.
What is the probability that the selected day will be Tuesday and the selected month will be January?

$F.\;\; \dfrac{1}{84} \\[5ex] G.\;\; \dfrac{1}{42} \\[5ex] H.\;\; \dfrac{1}{19} \\[5ex] J.\;\; \dfrac{2}{19} \\[5ex] K.\;\; \dfrac{19}{84} \\[5ex]$

$\underline{First\;\;Scenario:\;\; Days} \\[3ex] n(S) = n(Days) = 7 \\[3ex] n(Tuesday) = 1 \\[3ex] P(Tuesday) = \dfrac{n(Tuesday)}{n(Days)} = \dfrac{1}{7} \\[5ex] \underline{Second\;\;Scenario:\;\;Months} \\[3ex] n(S) = n(Months) = 12 \\[3ex] n(January) = 1 \\[3ex] P(January) = \dfrac{n(January)}{n(Months)} = \dfrac{1}{12} \\[5ex] P(1st\;\;Scenario\;\;AND\;\;2nd\;\;Scenario) \\[3ex] = \dfrac{1}{7} * \dfrac{1}{12} ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{1}{84}$
(127.)

(128.)

(129.) GCSE A school has 86 teachers.
42 are male and 44 are female.

$\dfrac{1}{3}$ of the male teachers have blue eyes.

$\dfrac{1}{4}$ of the female teachers have blue eyes.

(a) $\xi$ = teachers in the school
M = male teachers
B = teachers who have blue eyes

Complete the Venn diagram.

(b) One teacher who has blue eyes is chosen at random.
Work out the probability that the teacher is male.

$(a) \\[3ex] Male\;\;teachers\;\;with\;\;blue\;\;eyes \\[3ex] = n(M \cap B) \\[3ex] = \dfrac{1}{3} * 42 \\[5ex] = 14 \\[3ex] Male\;\;teachers = n(M) = 42 \\[3ex] n(Male\;\;teachers\;\;only) = 42 - 14 = 28 \\[3ex] Female\;\;teachers\;\;with\;\;blue\;\;eyes \\[3ex] = \dfrac{1}{4} * 44 \\[5ex] = 11 \\[3ex] Teachers\;\;with\;\;blue\;\;eyes \\[3ex] = n(blue\;\;eyes) \\[3ex] = 14 + 11 = 25 \\[3ex] n(Blue\;\;eyes\;\;only) = 25 - 14 = 11 \\[3ex] Let\;\;n(neither\;\;male\;\;nor\;\;blue\;\;eyes) = k \\[3ex] 28 + 14 + 11 + k = 86 \\[3ex] 53 + k = 86 \\[3ex] k = 86 - 53 \\[3ex] k = 33 \\[3ex]$ The Venn Diagram is:

$n(sample\;\;space) = n(B) = 25 \\[3ex] n(M \cap B) = 14 \\[3ex] P(M \cap B) = \dfrac{n(M \cap B)}{n(B)} = \dfrac{14}{25}$
(130.) ACT A weeklong summer camp is held in June for children in Grades 3 - 6
Parents and guardians who enrolled their children for camp by May 15 received a 20% discount off the regular enrollment fee for each child enrolled.
For each grade, the table below gives the number of children enrolled by May 15 as well as the regular enrollment fee per child.
The grade of any child is that child's grade in school as of May 15.

 Grade Enrollment by May 15 Regular enrollment fee 3 4 5 6 20 15 28 18 $350$400 $450$500

By May 15, Ms. Chen had enrolled her 2 children for camp.
One chid was in Grade 3, and the other was in Grade 4.
For each grade, the names of all the children enrolled by May 15 will be entered into a drawing for a free T-shirt.
For each grade, 1 name will be randomly drawn.
What is the probability that the names of both of Ms. Chen's children will be drawn?

$A.\;\; \dfrac{7}{60} \\[5ex] B.\;\; \dfrac{2}{81} \\[5ex] C.\;\; \dfrac{1}{300} \\[5ex] D.\;\; \dfrac{100}{2,187} \\[5ex] E.\;\; \dfrac{1}{3,240} \\[5ex]$

$n(Grade\;3) = 20 \\[3ex] n(1\;child\;\;in\;\;Grade\;3) = 1 \\[3ex] n(Grade\;4) = 15 \\[3ex] n(1\;child\;\;in\;\;Grade\;4) = 1 \\[3ex] P(1\;\;child\;\;Grade\;3\;\;AND\;\;1\;\;child\;\;Grade\;4) \\[3ex] = \dfrac{1}{20} * \dfrac{1}{15} \\[3ex] ...Multiplication\;\;Rule\;\;for\;\;Independent\;\;Events \\[3ex] = \dfrac{1}{300}$
(131.) WASSCE The number of green (G), red (R), white (W) and black (B) identical balls contained in a bag is as shown in the table.

 Balls $G$ $R$ $W$ $B$ Frequency $2$ $4$ $3$ $1$

If two balls are selected at random without replacement, find the probability that both balls are green.

$S = \{2G, 4R, 3W, 1B\} \\[3ex] n(S) = 2 + 4 + 3 + 1 = 10 \\[3ex] n(G) = 2 \\[3ex] \underline{Without\:\: Replacement - Dependent\:\: Events} \\[3ex] P(G \:\:AND\:\: G) = P(GG) \\[3ex] P(GG) = \dfrac{2}{10} * \dfrac{1}{9} ...Multiplication\:\: Rule \\[5ex] P(GG) = \dfrac{1}{5} * \dfrac{1}{9} \\[5ex] P(GG) = \dfrac{1 * 1}{5 * 9} \\[5ex] P(GG) = \dfrac{1}{45}$
(132.)

(133.) GCSE A and B are two events.
Some probabilities are shown on the Venn diagram.

Work out $P(A' \cup B)$

Please do this quickly...find the shaded area of $P(A' \cup B)$...as seen in Number (5.): Venn Diagram for Two Sets
As noted, we are concerned with those three shaded areas
But, first; let us find the probability of neither A nor B
The sum of all the probabilities is 1

$P(A \cup B)' = P(neither) \\[3ex] 0.3 + 0.15 + 0.35 + P(neither) = 1 \\[3ex] 0.8 + P(neither) = 1 \\[3ex] P(neither) = 1 - 0.8 \\[3ex] P(neither) = 0.2 \\[3ex]$ So, we need to add these shaded areas

$P(A' \cup B) \\[3ex] = 0.15 + 0.35 + 0.2 \\[3ex] = 0.7$
(134.) ACT "Snake-eyes" occur when you roll two 1's on a pair of regular, 6-sided dice numbered from 1 to 6
On any roll, what is the probability of rolling snake-eyes?

$F.\;\; \dfrac{1}{36} \\[5ex] G.\;\; \dfrac{1}{25} \\[5ex] H.\;\; \dfrac{1}{18} \\[5ex] J.\;\; \dfrac{1}{6} \\[5ex] K.\;\; \dfrac{1}{3} \\[5ex]$

The set containing (1, 1) is indicated in red

Sample Space for Two Rolls of a Fair Die
First Roll in the Column and Second Roll in the Row
$2nd\:\:Roll\:\:\rightarrow$
$1st\:\:Roll\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $\color{red}{(1, 1)}$ $(1, 2)$ $(1, 3)$ $(1, 4)$ $(1, 5)$ $(1, 6)$
$2$ $(2, 1)$ $(2, 2)$ $(2, 3)$ $(2, 4)$ $(2, 5)$ $(2, 6)$
$3$ $(3, 1)$ $(3, 2)$ $(3, 3)$ $(3, 4)$ $(3, 5)$ $(3, 6)$
$4$ $(4, 1)$ $(4, 2)$ $(4, 3)$ $(4, 4)$ $(4, 5)$ $(4, 6)$
$5$ $(5, 1)$ $(5, 2)$ $(5, 3)$ $(5, 4)$ $(5, 5)$ $(5, 6)$
$6$ $(6, 1)$ $(6, 2)$ $(6, 3)$ $(6, 4)$ $(6, 5)$ $(6, 6)$

$n(1, 1) = 1 \\[3ex] n(S) = 6 * 6 = 36 \\[3ex] P(1, 1) = \dfrac{n(1, 1)}{S} = \dfrac{1}{36}$
(135.)

(136.)

(137.) WASSCE-FM A fair coin is tossed 6 times.
Calculate the probability of obtaining at most three heads.

We can solve this question in at least two ways.
Use any approach you prefer.

First Method: Punnett Square
6 tosses of a fair coin
6 tosses = 4 tosses + 2 tosses
These will be in dark blue color
Sample Space for the Six Tosses of 1 Fair Coin

4 tosses in the Column and 2 tosses in the Row
$2\:\:Tosses\:\:\rightarrow$
$4\:\:Tosses\:\:\downarrow$
$HH$ $HT$ $TH$ $TT$
$HHHH$ $HHHHHH$ $HHHHHT$ $HHHHTH$ $HHHHTT$
$HHHT$ $HHHTHH$ $HHHTHT$ $HHHTTH$ $HHHTTT$
$HHTH$ $HHTHHH$ $HHTHHT$ $HHTHTH$ $HHTHTT$
$HHTT$ $HHTTHH$ $HHTTHT$ $HHTTTH$ $HHTTTT$
$HTHH$ $HTHHHH$ $HTHHHT$ $HTHHTH$ $HTHHTT$
$HTHT$ $HTHTHH$ $HTHTHT$ $HTHTTH$ $HTHTTT$
$HTTH$ $HTTHHH$ $HTTHHT$ $HTTHTH$ $HTTHTT$
$HTTT$ $HTTTHH$ $HTTTHT$ $HTTTTH$ $HTTTTT$
$THHH$ $THHHHH$ $THHHHT$ $THHHTH$ $THHHTT$
$THHT$ $THHTHH$ $THHTHT$ $THHTTH$ $THHTTT$
$THTH$ $THTHHH$ $THTHHT$ $THTHTH$ $THTHTT$
$THTT$ $THTTHH$ $THTTHT$ $THTTTH$ $THTTTT$
$TTHH$ $TTHHHH$ $TTHHHT$ $TTHHTH$ $TTHHTT$
$TTHT$ $TTHTHH$ $TTHTHT$ $TTHTTH$ $TTHTTT$
$TTTH$ $TTTHHH$ $TTTHHT$ $TTTHTH$ $TTTHTT$
$TTTT$ $TTTTHH$ $TTTTHT$ $TTTTTH$ $TTTTTT$

$n(S) = 16 * 4 = 64 \\[3ex] Let\;\;E = At\;\;most\;\;3H = dark\;blue\;\;color \\[3ex] n(E) = 42 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] = \dfrac{42}{64} \\[5ex] = 0.65625 \\[3ex]$ You can review the Second Method to solve the question:
Question (3.) using Binomial Distribution Formula
(138.) ACT In 3 fair coin tosses, what is the probability of obtaining exactly 2 tails?
(Note: In a fair coin toss, the 2 outcomes: heads and tails, are equally likely.)

$F.\;\; \dfrac{1}{3} \\[5ex] G.\;\; \dfrac{3}{8} \\[5ex] H.\;\; \dfrac{1}{2} \\[5ex] J.\;\; \dfrac{2}{3} \\[5ex] K.\;\; \dfrac{7}{8} \\[5ex]$

We can solve this question in at least two ways.
Use any approach you prefer.

First Approach: Punnett Square
The sets containing exactly 2 tails are indicated in red

Sample Space for Three Tosses of a Fair Coin
Two Tosses in the Column and One Toss in the Row
$One\:\:Toss\:\:\rightarrow$
$Two\:\:Tosses\:\:\downarrow$
$H$ $T$
$HH$ $HHH$ $HHT$
$HT$ $HTH$ $\color{red}{HTT}$
$TH$ $THH$ $\color{red}{THT}$
$TT$ $\color{red}{TTH}$ $TTT$

$n(exactly\;\;2\;\;tails) = 3 \\[3ex] n(S) = 4 * 2 = 8 \\[3ex] P(exactly\;\;2\;\;tails) \\[3ex] = \dfrac{n(exactly\;\;2\;\;tails)}{n(S)} \\[5ex] = \dfrac{3}{8} \\[5ex]$ You can review the Second Approach to solve the question:
Question (8.) using Binomial Distribution Formula
(139.) ACT Of the 16 cars on a rental-car lot, 6 are minivans, 7 are sedans, and 3 are hatchbacks.
Thalia will rent 3 of these cars, chosen at random, for business associates.
What is the probability that Thalia will rent 1 of each of the 3 types of cars?

$F.\;\; \dfrac{1}{3} \\[5ex] G.\;\; \dfrac{1}{16} \\[5ex] H.\;\; \dfrac{3}{16} \\[5ex] J.\;\; \dfrac{9}{40} \\[5ex] K.\;\; \dfrac{9}{80} \\[5ex]$

Please review the solution of this question using Combinatorics (Question 9.)
I suggest you use Combinatorics to solve it.
However, let us solve it using Probability.

Let:
minivan = M
sedans = E
hatchbacks = H

Things to note:
(a.) Renting 1 of each of the 3 cars means that the cars could be rented in any order
This means that Thalia could rent it in any of these order:
M AND E AND H
OR
M AND H AND E
OR
E AND M AND H
OR
E AND H AND M
OR
H AND M AND E
OR
H AND E AND M

(b.) The renting of each car is done without replacement because:
When Thalia rents 1 car of a type from the 16 cars, she will need to rent 1 car of another type from the remaining 15 cars, then she will rent 1 car of the other type from the remaining 14 cars.

$n(S) = 16 \\[3ex] n(M) = 6 \\[3ex] n(E) = 7 \\[3ex] n(H) = 3 \\[3ex] P(1\;\;car\;\;of\;\;each\;\;type) \\[3ex] = P(M-E-H) + P(M-H-E) + P(E-M-H) + P(E-H-M) + P(H-M-E) + P(H-E-M) \\[5ex] P(M-E-H) = \dfrac{6}{16} * \dfrac{7}{15} * \dfrac{3}{14} = \dfrac{3}{80} \\[5ex] P(M-H-E) = \dfrac{6}{16} * \dfrac{3}{15} * \dfrac{7}{14} = \dfrac{3}{80} \\[5ex] The\;\;rest\;\;of\;\;the\;\;probabilities\;\;are\;\;the\;\;same \\[3ex] But\;\;just\;\;to\;\;show\;\;you \\[3ex] P(E-M-H) = \dfrac{7}{16} * \dfrac{6}{15} * \dfrac{3}{14} = \dfrac{3}{80} \\[5ex] P(E-H-M) = \dfrac{7}{16} * \dfrac{3}{15} * \dfrac{6}{14} = \dfrac{3}{80} \\[5ex] P(H-M-E) = \dfrac{3}{16} * \dfrac{6}{15} * \dfrac{7}{14} = \dfrac{3}{80} \\[5ex] P(H-E-M) = \dfrac{3}{16} * \dfrac{7}{15} * \dfrac{6}{14} = \dfrac{3}{80} \\[5ex] P( = \dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} + dfrac{3}{80} \\[5ex] = \dfrac{3}{80} * 6 \\[5ex] = \dfrac{3}{40} * 3 \\[5ex] = \dfrac{9}{40}$
(140.)

(141.) ACT A certain committee is composed of 9 juniors and 11 seniors.
Two different members of the committee will be randomly selected for 2 different leadership roles.
Given that the 1st member who will be selected is a senior, what is the probability that the 2nd member who will be selected is a junior?

$A.\;\; \dfrac{9}{19} \\[5ex] B.\;\; \dfrac{9}{20} \\[5ex] C.\;\; \dfrac{10}{19} \\[5ex] D.\;\; \dfrac{10}{20} \\[5ex] E.\;\; \dfrac{11}{20} \\[5ex]$

When the 1st member (a senior) is selected, selecting the 2nd member is now dependent on the selection of the 1st member...Without Replacement condition

$Let: \\[3ex] junior = J \\[3ex] senior = E \\[3ex] n(J) = 9 \\[3ex] n(E) = 11 \\[3ex] n(S) = 9 + 11 = 20 \\[3ex] P(J | E) = \dfrac{P(J \cap E)}{P(E)}...Conditional\;\;Probability \\[5ex] P(J \cap E) = P(J) * P(E) \\[3ex] = \dfrac{9}{20} * \dfrac{11}{19}...Dependent\;\;Events \\[5ex] = \dfrac{99}{380} \\[5ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{11}{20} \\[5ex] \implies \\[3ex] P(J | E) \\[3ex] = \dfrac{99}{380} \div \dfrac{11}{20} \\[5ex] = \dfrac{99}{380} * \dfrac{20}{11} \\[5ex] = \dfrac{9}{19}$
(142.) GCSE $\mathcal{E}$ = {odd numbers less than 30}
A = {3, 9, 15, 21, 27}
B = {5, 15, 25}

(a) Complete the Venn diagram to represent this information.

A number is chosen at random from the universal set, $\mathcal{E}$
(b) What is the probability that the number is in the set $A \cup B$?

$\mathcal{E} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\} \\[3ex] A = \{3, 9, 15, 21, 27\} \\[3ex] A' = \{1, 5, 7, 11, 13, 17, 19, 23, 25, 29\} \\[3ex] B = \{5, 15, 25\} \\[3ex] B' = \{1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29\} \\[3ex] A \cap B = \{15\} \\[3ex] A \cup B = \{3, 5, 9, 15, 21, 25, 27\} \\[3ex] (A \cup B)' = \{1, 7, 11, 13, 17, 19, 23, 29\} \\[3ex] \underline{A\;\;only} \\[3ex] A \cap B' = \{3, 9, 21, 27\} \\[3ex] \underline{B\;\;only} \\[3ex] B \cap A' = \{5, 25\} \\[3ex]$ (a) The Venn diagram is:

$(b) \\[3ex] A \cup B = \{3, 5, 9, 15, 21, 25, 27\} \\[3ex] n(A \cup B) = 7 \\[3ex] \mathcal{E} = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29\} \\[3ex] n(\mathcal{E}) = 15 \\[3ex] P(A \cup B) = \dfrac{n(A \cup B)}{n(\mathcal{E})} = \dfrac{7}{15}$
(143.)

(144.) ACT Sets A, B, and C are defined below.
A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
B = {2, 4, 6, 8}
C= {4, 8}
A number will be randomly selected from set A.
What is the probability that the selected number will be an element of set C and an element of set B?

$F.\;\; \dfrac{1}{9} \\[5ex] G.\;\; \dfrac{2}{9} \\[5ex] H.\;\; \dfrac{4}{9} \\[5ex] J.\;\; \dfrac{6}{9} \\[5ex] K.\;\; 1 \\[3ex]$ Before you attempt this question, please review Question (82.)
Did you notice anything?

$A = \{1, 2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] n(A) = 9 \\[3ex] B \cap C = \{4, 8\} \\[3ex] n(B \cap C) = 2 \\[3ex] P(B \cap C) \\[3ex] = \dfrac{n(B \cap C)}{n(A)} \\[5ex] = \dfrac{2}{9}$
(145.)

(146.)

(147.)

(148.) ACT A bag contains exactly 18 solid-colored buttons: 3 red, 5 blue, and 10 white.
What is the probability of randomly selecting 1 button that is NOT white?

$F.\;\; \dfrac{1}{18} \\[5ex] G.\;\; \dfrac{1}{8} \\[5ex] H.\;\; \dfrac{4}{9} \\[5ex] J.\;\; \dfrac{2}{3} \\[5ex] K.\;\; \dfrac{4}{5} \\[5ex]$ Before you attempt this question, please review Question (27.)
Did you notice anything?

We can solve this in at least two ways
Choose whatever approach you prefer

$Let: \\[3ex] red = R \\[3ex] blue = B \\[3ex] white = W \\[3ex] S = \{3R, 5B, 10W\} \\[3ex] n(S) = 3 + 5 + 10 = 18 \\[5ex] \underline{First\:\: Method: Quantitative\:\: Reasoning} \\[3ex] NOT\:\:White \implies R\:\:AND\:\:B \\[3ex] n(R) = 3 \\[3ex] n(B) = 5 \\[3ex] n(R\:\:AND\:\:B) = 3 + 5 = 8 \\[3ex] P(NOT\:\:White) = P(R\:\:AND\:\:B) = \dfrac{n(R\:\:AND\:\:B)}{n(S)} \\[5ex] P(NOT\:\:White) = \dfrac{8}{18} \\[5ex] P(NOT\:\:White) = \dfrac{4}{9} \\[5ex] \underline{Second\:\: Method: Complementary\:\:Rule} \\[3ex] NOT\:\:white = W' \\[3ex] P(White) + P(NOT\:\:White) = 1 ... Complementary\:\: Rule \\[3ex] P(W) + P(W') = 1 \\[3ex] n(W) = 10 \\[3ex] P(W) = \dfrac{n(W)}{n(S)} = \dfrac{10}{18} \\[5ex] P(W') = 1 - P(W) \\[3ex] P(W') = 1 - \dfrac{10}{18} \\[5ex] P(W') = \dfrac{18}{18} - \dfrac{10}{18} \\[5ex] P(W') = \dfrac{18 - 10}{18} \\[5ex] P(W') = \dfrac{8}{18} \\[5ex] P(W') = \dfrac{4}{9}$
(149.) ACT Data from a random sample of 335 car owners in a certain city are listed below.
The table indicates the number of owners in 3 age brackets (16 - 25, 26 - 45, 46 - 60) who own cars from 3 car companies (A, B, C) in this city.
Each owner in the sample owns only 1 car.

 Car companies Age (in years) A B C Total 16 - 25 26 - 45 46 - 60 16 54 65 24 48 23 40 53 12 80 155 100 Total 135 95 105 335

Two cars owners from this sample will be chosen at random.
Given that no owner is chosen twice, which of the following expressions gives the probability that both owners chosen will be from the same age bracket?

$F.\;\; \dfrac{80(79)}{335(334)} + \dfrac{155(154)}{335(334)} + \dfrac{100(99)}{335(334)} \\[5ex] G.\;\; \dfrac{135(134)}{335(334)} + \dfrac{95(94)}{335(334)} + \dfrac{105(104)}{335(334)} \\[5ex] H.\;\; \dfrac{80}{335} + \dfrac{155}{335} + \dfrac{100}{335} \\[5ex] J.\;\; \dfrac{80}{335}\left(\dfrac{155}{334}\right)\left(\dfrac{100}{333}\right) \\[5ex] K.\;\; \dfrac{1}{3}\left(\dfrac{1}{3}\right)\left(\dfrac{1}{3}\right) \\[5ex]$

Things to NOTE:
(a.) Choosing two owners from the same age bracket implies choosing:
1 owner from ages 16 - 25 AND 1 owner from ages 16 - 25
OR
1 owner from ages 26 - 45 AND 1 owner from ages 26 - 45
OR
1 owner from ages 46 - 60 AND 1 owner from ages 46 - 60

(b.) Because an owner cannot be chosen twice, this is a case of Without Replacement

$P(2\;\;owners\;\;from\;\;ages\;\;16 - 25) = \dfrac{80}{335} * \dfrac{79}{334} \\[5ex] P(2\;\;owners\;\;from\;\;ages\;\;26 - 45) = \dfrac{155}{335} * \dfrac{154}{334} \\[5ex] P(2\;\;owners\;\;from\;\;ages\;\;46 - 60) = \dfrac{100}{335} * \dfrac{99}{334} \\[5ex] \implies \\[3ex] P(2\;\;owners\;\;from\;\;same\;\;age\;\;bracket) \\[3ex] = \dfrac{80}{335} * \dfrac{79}{334} + \dfrac{155}{335} * \dfrac{154}{334} + \dfrac{100}{335} * \dfrac{99}{334}$
(150.)

(151.) GCSE The Venn diagram shows some information about 150 students.
$\xi$ = 150 students
C = students who study Chemistry
P = students who study Physics

The probability that a Physics student, chosen at random, also studies Chemistry is $\dfrac{5}{12}$

One of the 150 students is chosen at random.
Work out the probability that the student does not study either Chemistry or Physics

$n(sample\;\;space) = n(P) = x + 35...number\;\;of\;\;Physics\;\;students \\[3ex] n(P \cap C) = x \\[3ex] P(P \cap C) = \dfrac{n(P \cap C)}{n(P)} = \dfrac{5}{12} \\[5ex] \implies \\[3ex] \dfrac{x}{x + 35} = \dfrac{5}{12} \\[5ex] 12x = 5(x + 35) \\[3ex] 12x = 5x + 175 \\[3ex] 12x - 5x = 175 \\[3ex] 7x = 175 \\[3ex] x = \dfrac{175}{7} \\[5ex] x = 25 \\[3ex] 47 + 35 + x + y = 150 \\[3ex] 82 + 25 + y = 150 \\[3ex] 107 + y = 150 \\[3ex] y = 150 - 107 \\[3ex] y = 43 \\[3ex] n(neither) = n(P \cup C)' = y = 43 \\[3ex] n(sample\;\;space) = n(\xi) = 150 \\[3ex] P(P \cup C)' = \dfrac{n(P \cup C)'}{n(\xi)} \\[5ex] = \dfrac{y}{150} \\[5ex] = \dfrac{43}{150}$
(152.) ACT The probabilities that each of 2 independent events will occur are given in the table below.

Event Probability
A
$0.20$
B
$0.40$

What is the probability that both Events A and B will occur - that is, $P(A\:\:and\:\:B)$ ?

$F.\:\: 0.08 \\[3ex] G.\:\: 0.20 \\[3ex] H.\:\: 0.30 \\[3ex] J.\:\: 0.50 \\[3ex] K.\:\: 0.60 \\[3ex]$

$P(A\:\:\:AND\:\:\:B) = P(A) * P(B) \\[3ex] ...Multiplication\:\:Rule\:\:for\:\:Independent\:\:Events \\[3ex] P(A\:\:\:AND\:\:\:B) = 0.20(0.40) \\[3ex] P(A\:\:\:AND\:\:\:B) = 0.08$
(153.)

(154.) JAMB

The Venn diagram below shows the number of students offering Music and History in a class of 80 students.
If a student is picked at random from the class, what is the probability that he offers Music only?

$A.\;\; 0.13 \\[3ex] B.\;\; 0.25 \\[3ex] C.\;\; 0.38 \\[3ex] D.\;\; 0.50 \\[3ex]$

$30 - x + 40 - x + x + 20 = 80 \\[3ex] 90 - x = 80 \\[3ex] 90 - 80 = x \\[3ex] 10 = x \\[3ex] x = 10 \\[3ex] n(Music\;\;only) \\[3ex] = 30 - x \\[3ex] = 30 - 10 \\[3ex] = 20 \\[3ex] P(Music\;\;only) = \dfrac{n(Music\;\;only)}{n(\mu)} = \dfrac{20}{80} = \dfrac{1}{4} = 0.25$
(155.) GCSE $\varepsilon$ = {2, 3, 4, 5, 6, 7, 8, 9}
P = {even numbers}
Q = {numbers divisible by 3}

(a) Complete the Venn diagram below.

(b) A number is chosen at random from the numbers 2 to 9
What is the probability that the number chosen is odd and not divisible by 3?

$\varepsilon = \{2, 3, 4, 5, 6, 7, 8, 9\} \\[3ex] P = \{2, 4, 6, 8\} \\[3ex] Q = \{3, 6, 9\} \\[3ex] P \cap Q = \{6\} \\[3ex] P \cup Q = \{2, 3, 4, 6, 8, 9\} \\[3ex] (P \cup Q)' = \{5, 7\} \\[3ex]$ (a)
The Venn diagram is:

(b)
Let the set: R = {odd and not divisible by 3}

$R = \{5, 7\} \\[3ex] n(R) = 2 \\[3ex] n(\varepsilon) = 8 \\[3ex] P(R) = \dfrac{n(R)}{n(\varepsilon)} = \dfrac{2}{8} = \dfrac{1}{4}$
(156.) ACT If it rains in Franklin City on a particular day, the probability that it will rain there the following day is 0.70
If it does not rain in Franklin City on a particular day, the probability that it will rain there the following day is 0.10
Given that it rained in Franklin City on Monday, what is the probability that it will NOT rain in Franklin City on Tuesday of the same week?

$F.\:\: 0.10 \\[3ex] G.\:\: 0.30 \\[3ex] H.\:\: 0.60 \\[3ex] J.\:\: 0.70 \\[3ex] K.\:\: 0.90 \\[3ex]$

It rained in Franklin City on Monday

The probability that it will rain there on Tuesday (the following day) is $0.70$

Let $T$ be the event that it will rain on Tuesday

This implies that $T'$ is the event that it will not rain on Tuesday... this is what we are asked to find

$P(T) = 0.70 \\[3ex] P(T) + P(T') = 1 ...Complementary\:\:Rule \\[3ex] P(T') = 1 - P(T) \\[3ex] P(T') = 1 - 0.70 \\[3ex] P(T') = 0.30$
(157.) GCSE In the Venn diagram
$\xi$ represents 31 students in a class
C is students who have a cat
D is students who have a dog

(a) One student from the class is picked at random.
Work out the probability that the student has a dog.

(b) One of the students who has a cat is picked at random.
Work out the probability that this student has a dog.

$6 + 2x + 5 + (x + 2) = 31 \\[3ex] 11 + 2x + x + 2 = 31 \\[3ex] 3x + 13 = 31 \\[3ex] 3x = 31 - 13 \\[3ex] 3x = 18 \\[3ex] x = \dfrac{18}{3} \\[5ex] x = 6 \\[3ex] (a) \\[3ex] n(sample\;\;space) = n(\xi) = 31 \\[3ex] n(event\;\;space) = n(D) \\[3ex] = 2x + 5 \\[3ex] = 2(6) + 5 \\[3ex] = 12 + 5 \\[3ex] = 17 \\[3ex] P(D) = \dfrac{n(D)}{n(\xi)} = \dfrac{17}{31} \\[5ex] (b) \\[3ex] n(sample\;\;space) = n(C) = 6 + 5 = 11 \\[3ex] n(event\;\;space) = n(C \cap D) = 5 \\[3ex] P(C \cap D) = \dfrac{n(C \cap D)}{n(C)} = \dfrac{5}{11}$
(158.) ACT A carnival game is played using an open box with a rectangular bottom measuring 6 inches by 13 inches.
A square with side lengths of 4 inches is painted on the bottom of the box.
The game is played by dropping a small bead into the open box.
If the bead comes to rest in the painted square, the player wins a prize.
Assuming a bead dropped into the box comes to rest at a random spot on the bottom of the box, which of the following is closest to the probability that the bead comes to rest in the painted square?

$A.\;\; 0.05 \\[3ex] B.\;\; 0.10 \\[3ex] C.\;\; 0.21 \\[3ex] D.\;\; 0.31 \\[3ex] E.\;\; 0.67 \\[3ex]$

$Area\;\;of\;\;rectangular\;\;box = 6(13) = 78\;square\;\;inches \\[3ex] Area\;\;of\;\;painted\;\;square = 4(4) = 16\;square\;\;inches \\[3ex] P(bead\;\;falls\;\;on\;\;painted\;\;square) \\[3ex] = \dfrac{16}{78} \\[5ex] = 0.2051282051 \\[3ex] \approx 0.21$
(159.)

(160.) GCSE Thirty shoppers were asked which of apples, oranges and bananas, they regularly liked to eat.
Of these 30 shoppers
• 2 liked all three fruits,
• 5 liked oranges but did not like apples or bananas,
• 6 liked both oranges and bananas,
• 17 liked apples,
• 10 liked bananas.
(a) Extra information from the 30 shoppers has been shown in the Venn diagram below.
Complete the Venn diagram using the information given above.

(b) One shopper is chosen at random.
Find the probability that this shopper does not like any of the three fruits.

(c) One shopper is chosen at random from those who like apples.
Find the probability that this shopper does not like oranges.

$\underline{Given:\;\;in\;\;English} \\[3ex] n(\varepsilon) = 30 \\[3ex] n(all\;\;three\;\;fruits) = 2 \\[3ex] n(oranges\;\;but\;\;not\;\;apples\;\;or\;\;bananas) = 5 \\[3ex] n(oranges\;\;and\;\;bananas) = 6 \\[3ex] n(apples) = 17 \\[3ex] n(bananas) = 10 \\[3ex] \underline{Given:\;\;in\;\;Diagram} \\[3ex] n(apples\;\;and\;\;oranges\;\;but\;\;not\;\;bananas) = 8 \\[3ex] n(apples\;\;and\;\;bananas\;\;but\;\;not\;\;oranges) = 1 \\[3ex] \underline{Determine} \\[3ex] n(oranges\;\;and\;\;bananas\;\;but\;\;not\;\;apples) \\[3ex] = 6 - 2 \\[3ex] = 4 \\[3ex] n(apples\;\;but\;\;not\;\;oranges\;\;or\;\;bananas) \\[3ex] = 17 - (1 + 8 + 2) \\[3ex] = 17 - 11 \\[3ex] = 6 \\[3ex] n(bananas\;\;but\;\;not\;\;apples\;\;or\;\;oranges) \\[3ex] = 10 - (1 + 2 + 4) \\[3ex] = 10 - 7 \\[3ex] = 3 \\[3ex] n(neither\;\;apples\;\;nor\;\;oranges\;\;nor\;\;bananas) = p \\[3ex] 6 + 5 + 3 + 1 + 8 + 4 + 2 + p = 30 \\[3ex] 29 + p = 30 \\[3ex] p = 30 - 29 \\[3ex] p = 1 \\[3ex]$ (a) The Venn diagram is:

$(b) \\[3ex] n(neither\;\;fruit) = p = 1 \\[3ex] n(\varepsilon) = 30 \\[3ex] P(neither\;\;fruit) = \dfrac{n(neither\;\;fruit)}{n(\varepsilon)} = \dfrac{1}{30} \\[5ex] (c) \\[3ex] \underline{From\;\;Apples} \\[3ex] n(apples) = 17 \\[3ex] n(apples\;\;but\;\;not\;\;oranges) = 6 + 1 = 7 \\[3ex] P(apples\;\;but\;\;not\;\;oranges) \\[3ex] = \dfrac{n(apples\;\;but\;\;not\;\;oranges)}{n(apples)} \\[5ex] = \dfrac{7}{17}$

(161.) WASSCE Out of 120 customers in a shop, 45 bought bags and shoes.
If all the customers bought either bags or shoes and 11 more customers bought shoes than bags:

(a) Illustrate this information in a diagram

(b) Find the number of customers who bought shoes

(c) Calculate the probability that a customer selected at random bought bags

$n(\xi) = 120 \\[3ex] Let: \\[3ex] bags = B \\[3ex] shoes = H \\[3ex] n(B \cap H) = n(both) = 45 \\[3ex] n(B) = n(bags) = b \\[3ex] \rightarrow n(H) = n(shoes) = 11 + b \\[3ex] n(bags\;\;only) = b - 45 \\[3ex] n(shoes\;\;only) = (11 + b) - 45 = 11 + b - 45 = b - 34 \\[3ex]$ (a)
The Venn diagram is:

$(b - 45) + (b - 34) + 45 = 120 \\[3ex] b - 45 + b - 34 + 45 = 120 \\[3ex] 2b = 120 + 34 \\[3ex] 2b = 154 \\[3ex] b = \dfrac{154}{2} \\[5ex] b = 77 \\[3ex] (b) \\[3ex] n(shoes) = n(H) \\[3ex] = (b - 34) + 45 \\[3ex] = b - 34 + 45 \\[3ex] = 77 - 34 + 45 \\[3ex] = 88 \\[3ex]$ 88 customers bought shoes

$(c) \\[3ex] n(bags) = n(B) \\[3ex] = (b - 45) + 45 \\[3ex] = 77 - 45 + 45 \\[3ex] = 77 \\[3ex] n(\xi) = 120 \\[3ex] P(bags) = \dfrac{n(bags)}{n\xi} = \dfrac{77}{120}$
(162.) ACT The probability that Event A will occur is 0.2
The probability that Event B will occur is 0.6
Given that Events A and B are mutually exclusive, what is the probability that Event A OR Event B will occur?

$F.\:\: 0.12 \\[3ex] G.\:\: 0.2 \\[3ex] H.\:\: 0.3 \\[3ex] J.\:\: 0.4 \\[3ex] K.\:\: 0.8 \\[3ex]$

$P(A\:\:\:OR\:\:\:B) = P(A) + P(B) \\[3ex] ...Addition\:\:Rule\:\:for\:\:Mutually\:\:Exclusive\:\:Events \\[3ex] P(A\:\:\:OR\:\:\:B) = P(A) + P(B) \\[3ex] P(A\:\:\:OR\:\:\:B) = 0.2 + 0.6 \\[3ex] P(A\:\:\:OR\:\:\:B) = 0.8$
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