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

probability

Welcome to Probability

I greet you this day,
First: read the notes. Second: view the videos. Third: solve the questions/solved examples. Fourth: check your solutions with my thoroughly-explained solutions.
Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome. You may contact me.
If you are my student, please do not contact me here. Contact me via the school's system. Thank you for visiting!!!

Samuel Dominic Chukwuemeka (Samdom For Peace) B.Eng., A.A.T, M.Ed., M.S

Objectives

Students will:

(1.) Discuss the meaning of probability.
(2.) Determine the probability of an event.
(3.) Compute the probabilities of events using the classical method.
(4.) Compute the probabilities of events using the empirical method.
(5.) Discuss scenarios related to probability including coins, dice, and cards among others.
(6.) Discuss the role of probability in life decisions including marriage among others.
(7.) Illustrate the sample spaces of events using probability tree diagrams.
(8.) Illustrate the sample spaces of events using Punnett squares.
(9.) Interpret the probabilities of events.
(10.) Determine the odds in favor of an event.
(11.) Determine the odds against an event.
(12.) Discuss the types of events.
(13.) Apply the rules of probability.
(14.) Solve applied problems involving probability.
(15.) Discuss conditional probability.
(16.) Solve applied problems involving conditional probability.
(17.) Discuss Bayes' theorem.

Mathematicians who did extensive work on Probability

(1.) Gregor Mendel - Mendelian Laws of Genetics - Roman Catholic Monk - Austrian
(2.) Reverend Thomas Bayes - Bayes Theorem/Conditional Probability - British
(3.) Reginald Punnett - Punnett Square - British
(4.) Girolamo Cardano - Italian
(5.) Pafnuty Chebyshev - Chebyshev's Theorem - Russian
(6.) Andrey Markov - Markov Chain - Russian
(7.) Andrey Kolmogorov - Russian
(8.) Pierre de Fermat - France
(9.) Pierre Simon de Laplace - France
(10.) Blaise Pascal - France
(11.) Honorable Mention - Evang. Samuel Dominic Chukwuemeka (SamDom For Peace) - The Humble Teacher

Okay, great people - please let me know whether I taught this topic well.
Did you learn anything from me? ☺☺☺

Scenarios Related to Probability / Vocabulary Words

Bring it to English Language: probably, chance, odds, odds of winning, odds of losing, placing bets, gambling, sweepstakes, raffle tickets, door prizes, cards, coins, dice, result, success, failure, events, set, dependent, independent, possible, impossible, odds, odds in favor, odds against,

Bring it to Science: phenomenon, random, experiment, outcome, theoretical, classical, empirical, theorem, rule, law,

Bring it to Mathematics: probability, sample space, event space, mutually exclusive, mutually inclusive, inclusive, with replacement, without replacement, addition law, multiplication law, inequality

Check for Prior Knowledge
Ask students to list all the ways/terms/scenarios in which they have used the word

(1.) Religion/Christianity: The probability that JESUS CHRIST will come again to judge all mankind is $1$ ($100\%$).

(2.) Meteorology: The National Weather Service predicted a $70\%$ chance of precipitation in Peachtree City, GA on the Saturday night of $3^{rd}$ February, $2018$

(3.) Gambling/Gaming: What are the odds in favor of winning the Powerball lottery?

(4.) Gambling/Gaming: What are the odds against winning the Mega Milions lottery?

(5.) Biology: The probability of seeing a pregnant man is $0$ $(0\%)$.

(6.) Statistics: It is generally known that one with a higher degree will earn more than another with a lower degree or no degree at all.
According to the Bureau of Labor Statistics, one of the reasons for varying wages is Credentials.
What is the probability that a randomly selected resident of the State of New Hampshire earns more than the minimum income for middle class according to CNBC given that the person has an advanced degree?
Did you notice the underlined, "given that"? What is the probability that an event will occur given that a prior event occurred?
This is known as Conditional Probability.

(7.) Criminology: Have you heard of a: true positive? true negative? false positive? false negative?
Is it possible for someone who is drug-free to test positive for drug-use? - Of course, it is possible.
Remember we are humans. Science helps us to avoid mistakes. With the advancement of science and technology, the "probability" of making those mistakes are small.
But, we still make mistakes because we are humans. We have limitations. Do you see another reason of using "probability"?
Is it also possible for someone who uses drugs to test negative for drug-use? - Of course, it is also possible.

and much more scenarios ...

Introduction

Let us define and explain several terms that we shall use in this topic.

Deterministic Phenomenon: is a phenomenon that you can predict the outcome based on the information you know.
For example: Assume you went to the Walmart store at a certain location to buy oranges.
Say the oranges were marked at $\$2.00$ per pound.
If you want to purchase $6$ pounds of oranges, how much would you pay?

Random Phenomenon: is a phenomenon that you cannot predict the outcome.
These include: the toss of a fair die, the pick of the winning numbers of a lottery ticket, and the toss of a fair coin among others.

Random Experiment: is the observation or measurement of a random phenomenon.
For example: forecasting the probability of earthquakes by observing the periods of occurrence, etc.

Outcome: is the result of an experiment.

Success: is a favorable outcome.
Assume you want to get a head whenever you toss a coin.
If you toss a coin and get a head, then it is success.
If you toss a coin and get a tail, then it is failure.

Failure: is an unfavorable outcome.

Sample Space: is the set of all possible outcomes of a random experiment.
It is generally denoted by $S$

Event Space: is a subset of the sample space which is also a set of some of all outcomes.
It is the set of the outcomes we want to find.
It is generally denoted by $E$

Events: We have five main types of events in Probability.
(1.) Dependent Events
(2.) Independent Events
(3.) Mutually Exclusive Events
(4.) Mutually Inclusive Events
(5.) Complementary Events

Please click on the "Events" link on the Left Hand Side ($LHS$) for further explanations.

Probability: of an event is the measure of the likelihood of the event.
OR
Probability: of an event is the measure of the success of the event.

The probability of an event can be expressed as a fraction, decimal, or percent.
$20\%$ chance or rain means that the probability that rain will fall is:
$20\%$ OR $\dfrac{20}{100} = \dfrac{1}{5}$ OR $\dfrac{20}{100} = 0.2$

So, $P(raining) = \dfrac{1}{5} \:\:or\:\: 0.2 \:\:or\:\: 20\%$


The Inequality Range: is $0 \le P(E) \le 1$

The probability of an impossible event (an event that will never happen) is $0$
The probability of a sure event (an event that must happen) is $1$
Therefore, the probability of any event is from $0$ through $1$
$0$ and $1$ are the two extremes for the probability of an event.
This means that if you are asked to calculate the probability of any event; if your answer is less than $0$ or greater than $1$, then it is incorrect.

We can classify Probability as:
(1.) Theoretical or Classical Probability
(2.) Empirical or Experimental Probability

Theoretical or Classical Probability: is the probability of an event in which each outcome of the event has an equal likelihood of occurrence.
The probability of an event, say $E$ is generally denoted by $P(E)$

$ P(E) = \dfrac{number\:\: of\:\: required\:\: outcomes}{number\:\: of\:\: possible\:\: outcomes} \\[5ex] \rightarrow P(E) = \dfrac{cardinality\:\: of\:\: event\:\: space}{cardinality\:\: of\:\: sample\:\: space} \\[5ex] \rightarrow P(E) = \dfrac{n(E)}{n(S)} $

Empirical or Experimental Probability: is the probability of an event in which each outcome of the event does not have an equal likelihood of occurrence.
It is based on short-run relative frequencies.
Experiments are needed.
The experiments used to produce empirical probabilities are known as simulations.

$ P(E) = \dfrac{number\:\: of\:\: times\:\: the\:\: event\:\: occurred}{number\:\: of\:\: times\:\: the\:\: experiment\:\: was\:\: performed} $

Simulations: are the experiments used to produce empirical probabilities.

A Fair Die: is a die in which all the outcomes are equally likely.

A Loaded Die: is a die in a certain outcome is more likely.

Tests
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.

Independent Events

Two events are independent if the probability of one event does not affect the probability of the other event.
For example: Say a bag contains $5$ red marbles, $3$ green marbles, and $4$ blue marbles.
Let:
$ Red = R \\[3ex] Green = G \\[3ex] Blue = B \\[3ex] n(R) = 5 \\[3ex] n(G) = 3 \\[3ex] n(B) = 4 \\[3ex] S = \{5R, 3G, 4B\} \\[3ex] n(S) = 5 + 3 + 4 = 12 \\[3ex] $ Event A: Samuel draws a marble from the bag.
It is a blue marble.
He does not like blue marbles.
So, he puts the marble back into the bag ... "With Replacement" condition

$ P(B) = \dfrac{n(B)}{n(S)} = \dfrac{4}{12} = \dfrac{1}{3} \\[5ex] $ Event B: Samuel draws another marble from the bag.
It is a green marble.

$ P(G) = \dfrac{n(G)}{n(S)} = \dfrac{3}{12} = \dfrac{1}{4} \\[5ex] $ We see that Events $A$ and $B$ are independent.
The probability of event $B$ occurring is not affected in any way by the probability of the occurrence of event $A$.
This is because of the With Replacement condition.
Samuel put that first pick back into the bag. So, all the marbles are complete prior to the second pick.

Dependent Events

Two events are dependent if the probability of one event affects the probability of the other event.
For example: Say a bag contains $5$ red marbles, $3$ green marbles, and $4$ blue marbles.
Let:

$ Red = R \\[3ex] Green = G \\[3ex] Blue = B \\[3ex] n(R) = 5 \\[3ex] n(G) = 3 \\[3ex] n(B) = 4 \\[3ex] S = \{5R, 3G, 4B\} \\[3ex] n(S) = 5 + 3 + 4 = 12 \\[3ex] $ Event C: Samuel draws a marble from the bag.
It is a blue marble.
He does not like blue marbles.
However, rather than putting it back into the bag, he throws the marble away.
So, the sample space has been affected. $n(S) = 11$ rather than $12$
This is because he threw the marble away ... "Without Replacement" condition

Event D: Samuel draws another marble from the bag.
It is a green marble.

$ P(B) = \dfrac{n(B)}{n(S)} = \dfrac{4}{12} = \dfrac{1}{3} \\[5ex] Without\:\: Replacement \\[3ex] S = \{5R, 3G, 3B\} \\[3ex] n(S) = 5 + 3 + 3 = 11 \\[3ex] P(G) = \dfrac{n(G)}{n(S)} = \dfrac{3}{11} \\[5ex] $ We see that the probability of picking a green marble was affected by the probability that a blue marble was picked earlier.
The event of picking the blue marble resulted in the decision to throw it away.
Throwing it away affects the sample space, and hence any other event/pick
Event $D$ was dependent on Event $C$
Events $C$ and $D$ are dependent events.
This is because of the Without Replacement condition.
Samuel did not replace the first pick. So, a marble was missing prior to the second pick.

Are there situations where we treat dependent events as independent events?
Yes.
It is known as the $\boldsymbol{5\%}$ Guideline for Treating Dependent Events as Independent Events
It states that: If the sample size is no more than (less than or equal to) $5\%$ of the population size, treat the selections as being independent.
$n \le 0.05N$

Mutually Exclusive Events

Two events are mutually exclusive if they cannot occur at the same time.
Those events are also said to be disjoint events
For example: In a single toss of a fair coin; the event of having a head, say Event $E$ and the event of having a tail, say Event $F$ at the same time are mutually exclusive.
In a single toss of a fair coin, one can either get a head or a tail, but not both.
Recall this topic in $XOR$ - Exclusive OR (Either or but not both)
For Mutually Exclusive Events, $P(E \cap F) = 0$

Mutually Inclusive Events

Two events are mutually inclusive if they can occur at the same time.
For example: In a single toss of a fair die; the event of having an odd number, say Event $G$ and the event of having a prime number, say Event $H$ at the same time are mutually inclusive.

Complementary Events

Two events are complementary if one event is the complement or prime of the other event.
The sum of the probabilities of an event and it's complement is $1$
The complement of an event, say $J$ is denoted by $J^c$ or $J'$
Let us review some examples.

Event $J$ Complement of Event $J$ is $J'$ or $J^c$
John will play the piano John will not play the piano
The couple will have at least 1 boy The couple will have no boy (all girls)

Odds of an Event

The Odds of an Event is the ratio of the number of ways the event can occur to the number of ways the event cannot occur.

$ Odds\:\:of\:\:an\:\:event = \dfrac{number\:\:of\:\:favorable\:\:outcomes}{number\:\:of\:\:unfavorable\:\:outcomes} \\[5ex] $ The odds of an event can be in favor of the event occurring, or against the event occurring.

$ Odds\:\:in\:\:favor = \dfrac{probability\:\:that\:\:event\:\:will\:\:occur}{probability\:\:that\:\:event\:\:will\:\:not\:\:occur} = \dfrac{P(E)}{P(E')} \\[5ex] P(E) = \dfrac{n(E)}{n(S)} \\[5ex] P(E') = \dfrac{n(E')}{n(S)} \\[5ex] \rightarrow Odds\:\:in\:\:favor = P(E) \div P(E') \\[3ex] Odds\:\:in\:\:favor = \dfrac{n(E)}{n(S)} \div \dfrac{n(E')}{n(S)} \\[5ex] Odds\:\:in\:\:favor = \dfrac{n(E)}{n(S)} * \dfrac{n(S)}{n(E')} \\[5ex] \rightarrow Odds\:\:in\:\:favor = \dfrac{n(E)}{n(E')} \\[5ex] Odds\:\:against = \dfrac{probability\:\:that\:\:event\:\:will\:\:not\:\:occur}{probability\:\:that\:\:event\:\:will\:\:occur} = \dfrac{P(E')}{P(E)} \\[5ex] \rightarrow Odds\:\:against = P(E') \div P(E) \\[3ex] Odds\:\:against = \dfrac{n(E')}{n(S)} \div \dfrac{n(E)}{n(S)} \\[5ex] Odds\:\:against = \dfrac{n(E')}{n(S)} * \dfrac{n(S)}{n(E)} \\[5ex] \rightarrow Odds\:\:against = \dfrac{n(E')}{n(E)} $

Usual and Unusual Events

In general; if the probability of an event is at least $5\%$ ($0.05$ or more), the event is usual.

In general (but depending on the situation); if the probability of an event is less than $5\%$ (less than $0.05$), the event is unusual.
What situation does this depend?
For example: In a life and death situation where a defendant is to be sentenced if guilty or released if innocent; the jurors need to be $\underline{100\%}$ certain of any decision they make. If the jurors are $96\%$ certain that the defendant is guilty, this means that the probability that the defendant is not guilty is $4\%$ (less than $5\%$). According to the $5\%$ rule, this implies that it is not usual that the defendant is not guilty.
But, what if the defendant is innocent? In this case that involves life or death, the event is not unusual.
LIFE IS PRECIOUS!!!

Multiplication Rule

Multiplication Rule goes with AND
Say we have two events: Event $A$ and Event $B$; what is the probability that both Event $A$ and Event $B$ occur?
Notice the "and"

The Multiplication Rule of Probability states that: Given two events, say Event $A$ and Event $B$; the probability that both event $A$ AND event $B$ occur is equal to the product of the individual probabilities of their occurrences.
$P(A\:\:\:AND\:\:\:B) = P(A) * P(B|A)$
$P(A \cap B) = P(A) * P(B|A)$
$P(A\:\:\:AND\:\:\:B) = P(A \cap B)$
$P(B|A)$ is read as: the probability of event $B$ given event $A$

For Independent Events
$ P(B|A) = P(B) \\[3ex] \rightarrow P(A\:\:\:AND\:\:\:B) = P(A) * P(B) \\[3ex] $ For Dependent Events
$ P(B|A) = P(B|A) \\[3ex] \rightarrow P(A\:\:\:AND\:\:\:B) = P(A) * P(B|A) $

Addition Rule

Addition Rule goes with OR
Say we have two events: Event $A$ and Event $B$; what is the probability that either Event $A$ or Event $B$ occurs?
Notice the "or"

The Addition Rule of Probability states that: Given two events, say Event $A$ and Event $B$; the probability that event $A$ OR event $B$ occur is equal to the probability that event $A$ occurs plus the probability that event $B$ occurs minus the probability that both events $A$ and $B$ occur.

Recall:
Based on the Definition of the union of two sets say set $A$ and set $B$;
Do you see the relationship between Sets and Probability
We are using our knowledge of Sets on Probability

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Bring it to Probability
Divide each term by $n(S)$

$ \dfrac{n(A \cup B)}{n(S)} = \dfrac{n(A)}{n(S)} + \dfrac{n(B)}{n(S)} - \dfrac{n(A \cap B)}{n(S)} \\[5ex] P(A \cup B) = P(A) + P(B) - P(A \cap B) \\[3ex] P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - P(A\:\:\:AND\:\:\:B) \\[3ex] $ For Independent Events
$ P(B|A) = P(B) \\[3ex] \rightarrow P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - [P(A) * P(B)] \\[3ex] $ For Dependent Events
$ P(B|A) = P(B|A) \\[3ex] \rightarrow P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - [P(A) * P(B|A)] \\[3ex] $ For Mutually Exclusive Events (Disjoint Events)
$ P(A \cap B) = 0 \\[3ex] P(A\:\:\:OR\:\:\:B) = P(A) + P(B) - 0 \\[3ex] \rightarrow P(A\:\:\:OR\:\:\:B) = P(A) + P(B) $

Complementary Rule

Complementary Rule applies only to Complementary Events
Say we have an events: Event $A$
The complement of Event $A$ is $A'$

$ P(A) + P(A') = 1 \\[3ex] \rightarrow P(A') = 1 - P(A) $

Rare Event Rule

The Rare Event Rule states that if under a given assumption, the probability of a particular observed event is extremely small (less than $5\%$), then the assumption is probably not correct.

Dice

A fair die has faces numbered $1, 2, 3, 4, 5, 6$

Case 1: A fair die tossed one time

$ S = \{1, 2, 3, 4, 5, 6\} \\[3ex] n(S) = 6 $

Case 2: A fair die tossed two times (twice) OR Two fair dice tossed one time

First Table for Case 2: List Each Number
$2^{nd}\:\:Die\:\:\rightarrow$
$1^{st}\:\:Die\:\:\downarrow$
$1$ $2$ $3$ $4$ $5$ $6$
$1$ $1, 1$ $1, 2$ $1, 3$ $1, 4$ $1, 5$ $1, 6$
$2$ $2, 1$ $2, 2$ $2, 3$ $2, 4$ $2, 5$ $2, 6$
$3$ $3, 1$ $3, 2$ $3, 3$ $3, 4$ $3, 5$ $3, 6$
$4$ $4, 1$ $4, 2$ $4, 3$ $4, 4$ $4, 5$ $4, 6$
$5$ $5, 1$ $5, 2$ $5, 3$ $5, 4$ $5, 5$ $5, 6$
$6$ $6, 1$ $6, 2$ $6, 3$ $6, 4$ $6, 5$ $6, 6$

$n(S) = 36$

Second Table for Case 2: Sum of Numbers
$(+)$ $1$ $2$ $3$ $4$ $5$ $6$
$1$ $2$ $3$ $4$ $5$ $6$ $7$
$2$ $3$ $4$ $5$ $6$ $7$ $8$
$3$ $4$ $5$ $6$ $7$ $8$ $9$
$4$ $5$ $6$ $7$ $8$ $9$ $10$
$5$ $6$ $7$ $8$ $9$ $10$ $11$
$6$ $7$ $8$ $9$ $10$ $11$ $12$

$n(S) = 36$

Student: Why do we need to draw the two tables?
Teacher: Good question.
Depending on the question, we shall use either table.
Let us do some examples

(1.) WAEC - FM The faces of a fair die are numbered $1, 2, 3, 4, 5, 6$.
If the die is thrown twice, what is the probability of obtaining a total score of $6$?


The question asked for a total score of $6$ (sum of $6$)
We shall use Case 2: Second Table
Let $E$ be the event of obtaining a total score of $6$

$ n(S) = 36 \\[3ex] n(E) = 5 \\[3ex] P(E) = \dfrac{n(E)}{n(S)} = \dfrac{5}{36} $

Cards

A standard deck of playing cards consists of four suits.
There are thirteen cards in each suit.
Altogether, the standard deck of playing has fifty two cards.
$n(cards) = 4 * 13 = 52$
The suits are: Clubs, Diamonds, Hearts, and Spades
The cards in each suit are: Ace, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, $10$, Jack, Queen, and King
The Jack, Queen, and King are called face cards (picture cards) because they have a a face on them.
The Jack, Queen, and King are also called court.
The Diamonds and Hearts are red.
The Spades and Clubs are black.

$ Let\:\: Sample\:\: Space = S \\[3ex] Clubs = C \\[3ex] Diamonds = D \\[3ex] Hearts = H \\[3ex] Spades = SP \\[3ex] Ace = A \\[3ex] Jack = J \\[3ex] Queen = Q \\[3ex] King = K \\[3ex] Suits = SU \\[3ex] Face\:\: Card = F-C \\[3ex] Red = R \\[3ex] Black = B \\[3ex] Non-Face\:\: Card = N-F-C \\[3ex] Court = CO \\[3ex] Numbers\:\: remain\:\: as\:\: is \\[3ex] n(S) = 52 \\[3ex] n(C) = 13 \\[3ex] n(D) = 13 \\[3ex] n(H) = 13 \\[3ex] n(SP) = 13 \\[3ex] n(R) = 13 + 13 = 26 \\[3ex] n(B) = 13 + 13 = 26 \\[3ex] n(F-C) = 3(4) = 12 \\[3ex] n(N-F-C) = 10(4) = 40 \\[3ex] OR \\[3ex] n(N-F-C) = 52 - 12 = 40 \\[3ex] n(A) = 4 \\[3ex] n(2) = 4 \\[3ex] n(3) = 4 \\[3ex] n(4) = 4 \\[3ex] n(5) = 4 \\[3ex] n(6) = 4 \\[3ex] n(7) = 4 \\[3ex] n(8) = 4 \\[3ex] n(9) = 4 \\[3ex] n(10) = 4 \\[3ex] n(J) = 4 \\[3ex] n(Q) = 4 \\[3ex] n(K) = 4 \\[3ex] n(CO) = 12 \\[3ex] n(Q\:\:of\:\: C) = 1 \\[3ex] n(K\:\:of\:\: H) = 1 \\[3ex] n(J\:\:of\:\: D) = 1 \\[3ex] $ Example: Cards
A card is randomly drawn from a standard deck of cards.
Determine the probability of selecting:

$ (1.)\:\:a\:\: 10 \\[3ex] \color{red}{P(10) = \dfrac{n(10)}{n(S)} = \dfrac{4}{52} = \dfrac{1}{13}} \\[5ex] (2.)\:\: an\:\: Ace \\[3ex] \color{red}{P(A) = \dfrac{n(A)}{n(S)} = \dfrac{4}{52} = \dfrac{1}{13}} \\[5ex] (3.)\:\: a\:\: King \\[3ex] \color{red}{P(K) = \dfrac{n(K)}{n(S)} = \dfrac{4}{52} = \dfrac{1}{13}} \\[5ex] (4.)\:\: a\:\: Red\:\: card \\[3ex] \color{red}{P(R) = \dfrac{n(R)}{n(S)} = \dfrac{26}{52} = \dfrac{1}{2}} \\[5ex] (5.)\:\: a\:\: Face\:\: card\:(Picture\:\: card) \\[3ex] \color{red}{P(F-C) = \dfrac{n(F-C)}{n(S)} = \dfrac{12}{52} = \dfrac{3}{13}} \\[5ex] (6.)\:\: a\:\: Non-face\:\: card\:(Non-picture\:\: card) \\[3ex] \color{red}{P(N-F-C) = \dfrac{n(N-F-C)}{n(S)} = \dfrac{40}{52} = \dfrac{10}{13}} \\[5ex] (7.)\:\:a\:\: Diamond \\[3ex] \color{red}{P(D) = \dfrac{n(D)}{n(S)} = \dfrac{13}{52} = \dfrac{1}{4}} \\[5ex] (8.)\:\: Queen\:\:of\:\:Clubs \\[3ex] \color{red}{P(Q\:\:of\:\:C) = \dfrac{n(Q\:\:of\:\:C)}{n(S)} = \dfrac{1}{52}} \\[5ex] (9.)\:\: a\:\: heart\:\: and\:\: a\:\: spade \\[3ex] $ From the question: A card is drawn one time.
It is not possible to get a heart and a spade at the same time.
The two events: the event of drawing a heart and the event of drawing a spade in a single selection of a card, are disjoint.
Hence, the events are mutually exclusive.

$ \color{red}{P(H \cap SP) = 0}...Mutually\:\: Exclusive\:\: Events \\[3ex] (10.)\:\: a\:\: King\:\: or\:\: Queen \\[3ex] \color{red}{P(K \:\:or\:\: Q) = \dfrac{n(K)}{n(S)} + \dfrac{n(Q)}{n(S)} ...Addition\:\: Rule \\[3ex] = \dfrac{4}{52} + \dfrac{4}{52} \\[3ex] = \dfrac{4 + 4}{52} \\[3ex] = \dfrac{8}{52} \\[3ex] = \dfrac{2}{13}} \\[5ex] (11.)\:\: a\:\: Club\:\: or\:\: a\:\: Diamond \:\:or\:\: a\:\: Spade \\[3ex] \color{red}{P(C \:\:or\:\: D \:\:or\:\: SP) = \dfrac{n(C)}{n(S)} + \dfrac{n(D)}{n(S)} + \dfrac{n(SP)}{n(S)} ...Addition\:\: Rule \\[3ex] = \dfrac{13}{52} + \dfrac{13}{52} + \dfrac{13}{52} \\[3ex] = \dfrac{13 + 13 + 13}{52} = \dfrac{39}{52} \\[3ex] = \dfrac{3}{4}} \\[5ex] $ (12.) A card is randomly drawn from a standard deck of cards.
The card drawn is court. (Jack, King, or Queen).
What is the probability that the card drawn is a king?

In other words, what is the probability that the card drawn is a king, given that this card is court?
This is Conditional Probability

$ \color{red}{P(K|CO) = \dfrac{n(K)}{n(CO)} \\[5ex] = \dfrac{4}{12} \\[5ex] = \dfrac{1}{3}} $

Solved Examples

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

(1.) Probability in Biology A man with $AS$ genotype wants to marry a woman with $AS$ genotype.
They are asking for your advice.
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 a Punnett Square to illustrate the likely genotypes of their four children

$Man/Woman$ $A$ $S$
$A$ $AA$ $AS$
$S$ $AS$ $SS$

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 trait.


(2.) 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 $A$


$A$ = {Ruth Kings, Esther Proverbs}

$ n(A) = 2 \\[3ex] n(S) = 9 \\[3ex] P(A) = \dfrac{n(A)}{n(S)} \\[5ex] P(A) = \dfrac{2}{9} \\[5ex] $ $A'$ = {Joshua Judges, Samuel Daniel, Nehemiah Job, Isaiah Psalms, Jeremiah Amos, Ezekiel Joel, Micah Nahum}


(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.) 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?


$ 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 - 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} $


(6.) 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?


$ 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] P(3 \:\:OR\:\: R) = \dfrac{3 + 2 - 1}{10} \\[5ex] P(3 \:\:OR\:\: R) = \dfrac{4}{10} = \dfrac{2}{5} $


(7.) 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} $


(8.) 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\% $


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\% $


(9.) 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 $\dfrac{3}{5}$?


Let the number of additional red marbles = $r$

$ \underline{Old} \\[3ex] n(S) = 42 \\[3ex] n(Red) = 16 \\[3ex] \underline{New} \\[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 \\[3ex] $ $23$ red marbles need to be added so that the probability of drawing a red marble is $\dfrac{3}{5}$


(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.

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

Numbers in blue 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 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 $


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 $


(12.) 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.


(13.) 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.


(14.) 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 \\[3ex] 120\% = \dfrac{120}{100} = 1.2 - NO \\[3ex] $ 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 $1$


(15.) A single die is rolled one time.
Determine the probability of obtaining an odd number or a number less than $5$


We can solve this in two ways

$ \underline{First\:\: Method: Quantitative\:\: Reasoning} \\[3ex] S = \{1, 2, 3, 4, 5, 6\} \\[3ex] n(S) = 6 \\[3ex] odd\:\: numbers = 1, 3, 5 \\[3ex] numbers \lt 5 = 1, 2, 3, 4 \\[3ex] odd\:\: number\:\: OR \:\:number \lt 5 = 1, 2, 3, 4, 5 \\[3ex] n(odd\:\: number\:\: OR \:\:number \lt 5) = 5 \\[3ex] \therefore P(odd\:\: number\:\: OR \:\:number \lt 5) = \dfrac{5}{6} \\[5ex] $ Let Event $A$ be the event of obtaining an odd number
Event $B$ be the event of obtaining a number less than 5

$ \underline{Second\:\: Method: Rule} \\[3ex] P(A \:\:OR\:\: B) = P(A) + P(B) - P(A \:\:AND\:\: B) ... Addition\:\: Rule \\[3ex] S = \{1, 2, 3, 4, 5, 6\} \\[3ex] n(S) = 6 \\[3ex] odd\:\: numbers = 1, 3, 5 \\[3ex] n(odd\:\: numbers) = 3 \\[3ex] numbers \lt 5 = 1, 2, 3, 4 \\[3ex] n(numbers \lt 5) = 4 \\[3ex] odd\:\: number\:\: AND \:\:number \lt 5 = 1, 3 \\[3ex] n(odd\:\: number\:\: AND \:\:number \lt 5) = 2 \\[3ex] \rightarrow P(A \:\:OR\:\: B) = \dfrac{3}{6} + \dfrac{4}{6} - \dfrac{2}{6} = \dfrac{3 + 4 - 2}{6} = \dfrac{5}{6} $


(16.) WAEC - FM 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 $


(17.) 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} $


(18.) 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)} = \dfrac{35}{100} = \dfrac{7}{20} $


(19.) 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.


(20.) A multiple-choice test has five possible answers.
However, only one answer is correct for each question.
Determine the probability of:
(a.) Answering a question correctly
(b.) Answering a question incorrectly


$ 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} $

(21.) 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} $


(22.) 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} $


(23.) $11$ 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} $


(24.) 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?


$ n(S) = 24 \\[3ex] n(A's) = 8 \\[3ex] P(A's) = \dfrac{n(A's)}{n(S)} = \dfrac{8}{24} = \dfrac{1}{3} $


(25.) 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} $


(26.) 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} $


(27.) The table below shows the "Math Guys" and the "Art Guys" among high school students ($10^{th}, 11^{th}, \:\:and\:\: 12^{th} \:\:grade\:\:students$) in a certain school in the City of Truth or Consequences, New Mexico

$10^{th}$ Grade $11^{th}$ Grade $12^{th}$ 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


$ 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.)\:\: P(Math) = \dfrac{n(Math)}{n(S)} = \dfrac{733}{1890} \\[5ex] n(11^{th}) = 510 \\[3ex] (b.)\:\: P(11^{th}) = \dfrac{n(11^{th})}{n(S)} = \dfrac{510}{1890} = \dfrac{17}{63} \\[5ex] n(12^{th}) = 112 \\[3ex] P(12^{th}) = \dfrac{n(12^{th})}{n(S)} = \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] (c.)\:\: P(11^{th}\:\:\:OR\:\:\:12^{th}) = \dfrac{510}{1890} + \dfrac{112}{1890} - 0 = \dfrac{510 + 112}{1890} = \dfrac{622}{1890} = \dfrac{311}{945} \\[5ex] n(11^{th}\:\:\:AND\:\:\:Math) = 442 \\[3ex] (d.)\:\: P(11^{th}\:\:\:AND\:\:\:Math) = \dfrac{n(11^{th}\:\:\:AND\:\:\:Math)}{n(S)} = \dfrac{442}{1890} = \dfrac{221}{945} \\[5ex] P(11^{th}\:\:\:OR\:\:\:Math) = P(11^{th}) + P(Math) - P(11^{th}\:\:\:AND\:\:\:Math) ... Addition\:\:Rule \\[3ex] (e.)\:\: P(11^{th}\:\:\:OR\:\:\:Math) = \dfrac{510}{1890} + \dfrac{733}{1890} - \dfrac{442}{1890} = \dfrac{510 + 733 - 442}{1890} = \dfrac{267}{630} = \dfrac{89}{210} \\[5ex] (f.)\:\: P(Math\:\:from\:\:11^{th}) = \dfrac{n(11^{th}\:\:\:AND\:\:\:Math)}{n(11^{th})} = \dfrac{442}{510} = \dfrac{13}{15} \\[5ex] n(12^{th}\:\:\:AND\:\:\:Math) = 59 \\[3ex] (g.)\:\: P(Math\:\:from\:\:12^{th}) = \dfrac{n(12^{th}\:\:\:AND\:\:\:Math)}{n(12^{th})} = \dfrac{59}{112} \\[5ex] n(10^{th}) = 1268 \\[3ex] n(10^{th}\:\:\:AND\:\:\:Math) = 232 \\[3ex] (h.)\:\: P(Math\:\:from\:\:10^{th}) = \dfrac{n(10^{th}\:\:\:AND\:\:\:Math)}{n(10^{th})} = \dfrac{232}{1268} = \dfrac{58}{317} \\[5ex] P(Two\:\:Math\:\:from\:\:11^{th}) = P(Math\:\:from\:\:11^{th}) * P(Math\:\:from\:\:11^{th}) ... Multiplication\:\:Rule \\[5ex] (i.)\:\: P(Two\:\:Math\:\:from\:\:11^{th}) = \dfrac{13}{15} * \dfrac{13}{15} = \dfrac{13 * 13}{15 * 15} = \dfrac{169}{225} $


(28.) 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?


$ 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} $


(29.) 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} $


(30.) Samuel, Dominic, and Chukwuemeka work for SNHU.
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} $


(31.) 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} $


(32.) 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} $


(33.) ACT A bag contains $10$ pieces of favored 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?


$ 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) ... 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} $


(34.) 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} $


References


Chukwuemeka, S.D (2016, April 30). Samuel Chukwuemeka Tutorials - Math, Science, and Technology. Retrieved from https://www.chukwuemekasamuel.com
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