Word Problems on Applications of Combinatorics in Probability

Fundamental Counting Principle in Probability

We were not given the number of books.
Let the number of books be $p$
The number of ways of revieweing all the books is $p!$

For the first question:
When Book $C$ is reviewed first, the remaining number of books is $p - 1$
There is only $1$ way of reviewing Book $C$
When this is done, the remaining books can be reviewed in $(p - 1)!$ ways.
So, the number of ways of arranging the books to review such that Book $C$ is reviewed first is $1 * (p - 1)!$ ways...Fundamental Counting Principle.

$ \underline{Fundamental\;\;Counting\;\;Principle} \\[3ex] Book\;\;C\;\;to\;\;be\;\;reviewed\;\;first \\[3ex] n(C) = Number\;\;of\;\;ways = 1 * (p - 1)! \;\;ways \\[3ex] n(S) = total\;\;number\;\;of\;\;ways\;\;of\;\;reviewing\;\;all\;\;books \\[3ex] n(S) = p! \\[3ex] (a.) \\[3ex] P(Book\;\;C) = \dfrac{n(C)}{n(S)} \\[5ex] = \dfrac{1 * (p - 1)!}{p!} \\[5ex] = \dfrac{(p - 1)!}{p * (p - 1)!} \\[5ex] = \dfrac{1}{p} \\[5ex] $ For the second question:
After reviewing Book $C$, there is only $1$ way of reviewing Book $V$
The remaining books can then be reviewed in $(p - 2)!$ ways.

$ Book\;\;C\;\;to\;\;be\;\;reviewed\;\;first,\;\;Book\;\;V\;\;to\;\;be\;\;reviewed\;\;next \\[3ex] n(V) = Number\;\;of\;\;ways = 1 * 1 * (p - 2)! \;\;ways \\[3ex] n(S) = p! \\[3ex] (b.) \\[3ex] P(Book\;\;V) = \dfrac{n(V)}{n(S)} \\[5ex] = \dfrac{1 * 1 * (p - 2)!}{p!} \\[5ex] = \dfrac{(p - 2)!}{p * (p - 1) * (p - 2)!} \\[5ex] = \dfrac{1}{p(p - 1)} \\[5ex] $ For the third question:
The order of arrangement is not important.
But Jude chose to place Book $C$ and Book $V$ next to each other.
This implies that Book $C$ can be placed first AND Book $V$ next;
OR Book $V$ can be placed first AND Book $C$ next.
The number of ways of doing this is: $1 * (p - 1)! + 1 * (p - 1)!$ ways

$ 1 * (p - 1)! + 1 * (p - 1)! \\[3ex] = (p - 1)! + (p - 1)! \\[3ex] = 2(p - 1)! \\[3ex] (c.) \\[3ex] Book\;\;C\;\;and\;\;Book\;\;V\;\;placed\;\;next\;\;to\;\;each\;\;other \\[3ex] n(CV\;\;OR\;\;VC) = 2(p - 1)! \\[3ex] n(S) = p! \\[3ex] P(CV\;\;OR\;\;VC) = \dfrac{n(CV\;\;OR\;\;VC)}{n(S)} \\[5ex] = \dfrac{2(p - 1)!}{p!} \\[5ex] = \dfrac{2(p - 1)!}{p * (p - 1)!} \\[5ex] = \dfrac{2}{p} \\[5ex] $ For the fourth question:
Let us consider these three events.
Event $1$: Jude placed Book $C$ first
Event $2$: Jude placed Book $V$ last
Event $3$: Jude placed Book $C$ first and Book $V$ last

The number of ways of placing Book $C$ first is $1 * (p - 1)!$ ways.
This includes the events where Book $V$ is placed last.

The number of ways of placing Book $V$ last is also $1 * (p - 1)$ ways.
This also includes the events where Book $C$ is placed first.

The number of ways of placing Book $C$ first and Book $V$ last is $1 * 1 * (p - 2)$ ways

Therefore, for us to find the number of ways of placing Book $C$ first or Book $v$ last, we need to avoid double counts.
We have to use the Addition Rule for Mutually Inclusive Events
This states that:
The number of ways of placing Book $C$ first or Book $V$ last is equal to:
the number of ways of placing Book $C$ first plus
the number of ways of placing Book $V$ last minus
the number of ways of placing Book $C$ first and BOok $V$ last

$ (d.) \\[3ex] n(Book\;C\;\;is\;\;placed\;\;first) = 1 * (p - 1)! = (p - 1)! \\[3ex] n(Book\;V\;\;is\;\;placed\;\;last) = 1 * (p - 1)! = (p - 1)! \\[3ex] n(Book\;C\;\;is\;\;placed\;\;first\;\;AND\;\;Book\;\;V\;\;is\;\;placed\;\;last) = 1 * 1 * (p - 2)! = (p - 2)! \\[3ex] n(Book\;C\;\;is\;\;placed\;\;first\;\;OR\;\;Book\;\;V\;\;is\;\;placed\;\;last) \\[3ex] = (p - 1)! + (p - 1)! - (p - 2)! ...Addition\;\;Rule\;\;for\;\;Mutually\;\;Inclusive\;\;Events \\[3ex] n(S) = p!...cardinality\;\;of\;\;the\;\;sample\;\;space \\[3ex] P(Book\;C\;\;first\;\;OR\;\;Book\;\;V\;\;last) = \dfrac{n(Book\;C\;\;first\;\;OR\;\;Book\;\;V\;\;last)}{n(S)} \\[5ex] = \dfrac{(p - 1)! + (p - 1)! - (p - 2)!}{p!} \\[5ex] = \dfrac{(p - 1) * (p - 1 - 1)! + (p - 1) * (p - 1 - 1)! - (p - 2)!}{p * (p - 1) * (p - 2)!} \\[5ex] = \dfrac{(p - 1) * (p - 2)! + (p - 1) * (p - 2)! - (p - 2)!}{p * (p - 1) * (p - 2)!} \\[5ex] = \dfrac{(p - 2)![(p - 1) + (p - 1) - 1]}{p * (p - 1) * (p - 2)!} \\[5ex] = \dfrac{p - 1 + p - 1 - 1}{p(p - 1)} \\[5ex] = \dfrac{2p - 3}{p(p - 1)} $

Combination in Probability

$ 1\;\;marble\;\;is\;\;selected \\[3ex] The\;\;marble\;\;should\;\;be\;\;red \\[3ex] 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] n(P) = 35 \\[3ex] P(R) = \dfrac{C(4, 1)}{C(12, 1)} \\[5ex] C(n, r) = \dfrac{n!}{(n - r) r!} \\[5ex] C(4, 1) \\[3ex] = \dfrac{4!}{(4 - 1)! 1!} \\[5ex] = \dfrac{4 * 3!}{3! * 1!} \\[5ex] = 4 \\[3ex] C(12, 1) \\[3ex] = \dfrac{12!}{(12 - 1)! 1!} \\[5ex] = \dfrac{12 * 11!}{11! * 1!} \\[5ex] = 12 \\[3ex] \implies P(R) = \dfrac{4}{12} \\[5ex] P(R) = \dfrac{1}{3} $

Combination in Probability

$ C(n, r) = \dfrac{n!}{(n - r) r!} \\[5ex] Number\;\;in\;\;group = 8\;\;boys + 5\;\;girls = 13\;\;people \\[3ex] Number\;\;in\;\;committee = 4\;\;boys + 3\;\;girls = 7\;\;people \\[3ex] Number\;\;of\;\;ways\;\;of\;\;selecting\;\;7\;\;members\;\;from\;\;a\;\;group\;\;of\;\;13\;\;members \\[3ex] = C(13, 7) \\[3ex] = \dfrac{13!}{(13 - 7)! 7!} \\[5ex] = \dfrac{13 * 12 * 11 * 10 * 9 * 8 * 7!}{6! * 7!} \\[5ex] = \dfrac{13 * 12 * 11 * 10 * 9 * 8}{6 * 5 * 4 * 3 * 2 * 1} \\[5ex] = \dfrac{13 * 11 * 9 * 8}{6} \\[5ex] = \dfrac{13 * 11 * 3 * 8}{2} \\[5ex] = 13 * 11 * 3 * 4 \\[3ex] = 1716 \\[3ex] Number\;\;of\;\;ways\;\;of\;\;selecting\;\;4\;\;boys\;\;from\;\;8\;\;boys \\[3ex] = C(8, 4) \\[3ex] = \dfrac{8!}{(8 - 4)! * 4!} \\[5ex] = \dfrac{8 * 7 * 6 * 5 * 4!}{4! * 4 * 3 * 2 * 1} \\[5ex] = 2 * 7 * 5 \\[3ex] = 70 \\[3ex] Number\;\;of\;\;ways\;\;of\;\;selecting\;\;3\;\;girls\;\;from\;\;5\;\;girls \\[3ex] = C(5, 3) \\[3ex] = \dfrac{5 * 4 * 3!}{(5 - 3)! * 3!} \\[5ex] = \dfrac{5 * 4}{2!} \\[5ex] = \dfrac{5 * 4}{2 * 1} \\[5ex] = 5 * 2 \\[3ex] = 10 \\[3ex] Probability\;\;of\;\;selecting\;\;4\;\;boys\;\;from\;\;8\;\;boys \\[3ex] AND\;\;3\;\;girls\;\;from\;\;5\;\;girls \\[3ex] = \dfrac{C(8, 4) * C(5, 3)}{C(13, 7)} \\[5ex] = \dfrac{70 * 10}{1716} \\[5ex] = \dfrac{700}{1716} \\[5ex] = \dfrac{175}{429} \\[5ex] = 0.4079254079 $

Combination in Probability

$ 1\;\;marble\;\;is\;\;selected \\[3ex] The\;\;marble\;\;should\;\;be\;\;purple \\[3ex] Let\:\: purple = P \\[3ex] n(S) = 100 \\[3ex] n(P) = 35 \\[3ex] P(P) = \dfrac{C(35, 1)}{C(100, 1)} \\[5ex] C(n, r) = \dfrac{n!}{(n - r) r!} \\[5ex] C(35, 1) \\[3ex] = \dfrac{35!}{(35 - 1)! 1!} \\[5ex] = \dfrac{35 * 34!}{34! * 1!} \\[5ex] = 35 \\[3ex] C(100, 1) \\[3ex] = \dfrac{100!}{(100 - 1)! 1!} \\[5ex] = \dfrac{100 * 99!}{99! * 1!} \\[5ex] = 100 \\[3ex] \implies P(P) = \dfrac{35}{100} \\[5ex] P(P) = \dfrac{7}{20} $

Fundamental Counting Principle in Probability

Based on the Formulas listed in the Note:
Given: a certain number of people or items in a linear random order say $n$
The number of ways in which two people or two items must be close together is $2 * (n - 1) * (n - 2)!$ ways

For the first question:
Students $A$ and $C$ must always stand next to each other.
This can happen in two ways and can occur in six positions: Positions $1$ and $2$; $2$ and $3$; $3$ and $4$; $4$ and $5$; $5$ and $6$; and $6$ and $7$
So, there are $2 * 6 = 12$ different ways they can stand next to each other.
When this happens, the other $7 - 2 = 5$ students can stand in $5!$ ways
Hence, the number of ways that Students $A$ and $C$ must stand next to each other is $12 * 5!$ ways
Just as it is in the formula: $2 * (7 - 1) * (7 - 2)!$

The number of ways for the $7$ students to stand in a linear order = $7!$

$ (I.) \\[3ex] Let\;\;A\;\;be\;\;the\;\;event\;\;that\;\;Students\;\;A\;\;and\;\;C\;\;stand\;\;next\;\;to\;\;each\;\;other \\[3ex] n(A) = 2 * 6 * 5! = 12 * 120 = 1440 \\[3ex] n(S) = 7! = 5040 \\[3ex] P(A) = \dfrac{n(A)}{n(S)} \\[5ex] = \dfrac{1440}{5440} \\[5ex] = \dfrac{9}{34} \\[5ex] $ For the second question:
There is one way of standing Student $D$ in the rightmost position.
Once this is done, the rest of the students: $7 - 1 = 6$ can stand in $6!$ ways

$ (I.) \\[3ex] Let\;\;B\;\;be\;\;the\;\;event\;\;that\;\;Student\;\;D\;\;is\;\;in\;\;rightmost\;\;position \\[3ex] n(B) = 1 * 6! = 1 * 720 = 720 \\[3ex] n(S) = 7! = 5040 \\[3ex] P(B) = \dfrac{n(B)}{n(S)} \\[5ex] = \dfrac{720}{5440} \\[5ex] = \dfrac{9}{68} \\[5ex] $ For the third question:
Two events say: $A$ and $B$ are Independent if:

$ (III.) \\[3ex] P(A \cap B) = P(A) * P(B)...Multiplication\;\;Rule\;\;for\;\;Independent\;\;events \\[3ex] Let: \\[3ex] Event\;A:\;\; Students\;\;A\;\;and\;\;C\;\;stand\;\;next\;\;to\;\;each\;\;other \\[3ex] Event\;B:\;\; Student\;\;D\;\;is\;\;in\;\;rightmost\;\;position \\[3ex] Event\;(A \cap B):\;\;this\;\;order:\;\;...,\; ...,\; ...,\; ...,\; A,\;C,\;D \\[3ex] Number\;\;of\;\;ways\;\;for\;\;Event\;(A \cap B) = 1 * 1 * 5! = 120 \\[3ex] P(A \cap B) = \dfrac{n(A \cap B)}{n(S)} = \dfrac{120}{5040} = \dfrac{3}{126} \\[5ex] P(A) = \dfrac{9}{34} \\[5ex] P(B) = \dfrac{9}{68} \\[5ex] P(A) * P(B) = \dfrac{9}{34} * \dfrac{9}{68} = \dfrac{81}{2312} \\[5ex] \dfrac{3}{126} \ne \dfrac{81}{2312} \\[5ex] P(A \cap B) \ne P(A) * P(B) \\[3ex] $ Therefore, the events in $(I.)$ and $(II.)$ are not independent.

Poisson Distribution: Probability in Combinatorics

$ P = \dfrac{3^n e^{-3}}{n!} \\[5ex] n = 2 \\[3ex] e^{-3} \approx 0.05 \\[3ex] P = \dfrac{3^2 * 0.05}{2!} \\[5ex] P = \dfrac{9 * 0.05}{2 * 1} \\[5ex] P = \dfrac{0.45}{2} \\[5ex] P = 0.225 \\[3ex] P \approx 0.23 $

Combination in Probability

$ Number\;\;of\;\;jurors = 1 + 3 + 2 = 6 \\[3ex] C(n, r) = \dfrac{n!}{(n - r) r!} \\[5ex] Number\;\;of\;\;ways\;\;of\;\;selecting\;\;2\;\;jurors\;\;out\;\;of\;\;6\;\;jurors \\[3ex] = C(6, 2) \\[3ex] = \dfrac{6!}{(6 - 2)! 2!} \\[5ex] = \dfrac{6 * 5 * 4!}{4! * 2 * 1} \\[5ex] = 3 * 5 \\[3ex] = 15 \\[3ex] Number\;\;of\;\;ways\;\;of\;\;selecting\;\;1\;\;American\;\;from\;\;3\;\;Americans \\[3ex] = C(3, 1) \\[3ex] = \dfrac{3!}{(3 - 1)! * 1!} \\[5ex] = \dfrac{3 * 2!}{2! * 1} \\[5ex] = 3 \\[3ex] Number\;\;of\;\;ways\;\;of\;\;selecting\;\;1\;\;African\;\;from\;\;2\;\;Africans \\[3ex] = C(2, 1) \\[3ex] = \dfrac{2!}{(2 - 1)! * 1!} \\[5ex] = \dfrac{2 * 1!}{1! * 1} \\[5ex] = 2 \\[3ex] (a.) \\[3ex] Probability\;\;of\;\;selecting\;\;1\;\;American\;\;and\;\;1\;\;African \\[3ex] = \dfrac{C(3, 1) * C(2, 1)}{C(6, 2)} \\[5ex] = \dfrac{3 * 2}{15} \\[5ex] = \dfrac{2}{5} \\[5ex] (b.) \\[3ex] Number\;\;of\;\;ways\;\;of\;\;selecting\;\;1\;\;British\;\;from\;\;1\;\;British \\[3ex] = C(1, 1) \\[3ex] = \dfrac{1!}{(1 - 1)! * 1!} \\[5ex] = \dfrac{1!}{0! * 1} \\[5ex] = 1 \\[3ex] $ Selecting 2 jurors of different nationalities could be:
1 British from 1 British AND 1 American from 3 Americans
OR
1 American from 3 Americans AND 1 African from 2 Africans
OR
1 British from 1 British AND 1 African from 2 Africans
This means:

$ Number\;\;of\;\;ways\;\;of\;\;selecting\;\;2\;\;jurors\;\;from\;\;different\;\;nationalities \\[3ex] = C(1, 1) * C(3, 1)\;\;OR\;\;C(3, 1) * C(2, 1)\;\;OR\;\;C(1, 1) * C(2, 1) \\[3ex] = (1 * 3) + (3 * 2) + (1 * 2) \\[3ex] = 3 + 6 + 2 \\[3ex] = 11 \\[3ex] Probability\;\;of\;\;selecting\;\;2\;\;jurors\;\;from\;\;different\;\;nationalities \\[3ex] = \dfrac{11}{15} $

Thalia can rent minivan first, then sedan, then hatchback
Thalia can also rent sedan first, minivan second, and hatchback third
...and so on and so forth
The order does not matter. Hence it is:
Combination in Probability

$ C(n, r) = \dfrac{n!}{(n - r) r!} \\[5ex] Total\;\;number\;\;of\;\;ways = C(16, 3) \\[3ex] = \dfrac{16!}{(16 - 3)! * 3!} \\[5ex] = \dfrac{16 * 15 * 14 * 13!}{13! * 3 * 2 * 1} \\[5ex] = 16 * 5 * 7 \\[3ex] = 560 \\[3ex] Number\;\;of\;\;ways\;\;of\;\;Thalia\;\;renting\;\;1\;\;of\;\;6\;\;minivans \\[3ex] = C(6, 1) \\[3ex] = \dfrac{6!}{(6 - 1)! * 1!} \\[5ex] = \dfrac{6 * 5!}{5! * 1} \\[5ex] = 6 \\[3ex] Number\;\;of\;\;ways\;\;of\;\;Thalia\;\;renting\;\;1\;\;of\;\;7\;\;sedans \\[3ex] = C(7, 1) \\[3ex] = \dfrac{7!}{(7 - 1)! * 1!} \\[5ex] = \dfrac{7 * 6!}{6! * 1} \\[5ex] = 7 \\[3ex] Number\;\;of\;\;ways\;\;of\;\;Thalia\;\;renting\;\;1\;\;of\;\;3\;\;hatchbacks \\[3ex] = C(3, 1) \\[3ex] = \dfrac{3!}{(3 - 1)! * 1!} \\[5ex] = \dfrac{3 * 2!}{2! * 1} \\[5ex] = 3 \\[3ex] Probability\;\;of\;\;Thalia\;\;renting\;\;1\;\;of\;\;each\;\;of\;\;the\;\;3\;\;types\;\;of\;\;cars \\[3ex] = \dfrac{C(6, 1) * C(7, 1) * C(3, 1)}{C(16, 3)} \\[5ex] = \dfrac{6 * 7 * 3}{560} \\[5ex] = \dfrac{6 * 3}{80} \\[5ex] = \dfrac{3 * 3}{40} \\[5ex] = \dfrac{9}{40} $