Word Problems on Applications of Combinatorics in 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 a wrong answer. Use the functions in your TI-84 Plus or TI-Nspire to solve some of the questions in order to save
time.
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.
Use at least two methods whenever applicable.
State the case/type of combinatorics for each question.
Show all work.
(1.) Jude has a certain number of books to review.
He labeled each book as Book $A$, Book $B$, Book $C$ among others.
However, the order in which he reviews them is not important.
He chose to arrange these books horizontally on the desk.
Each book has an equal likelihood of being reviewed.
(a.) Determine the probability that Book $C$ is the first book to be reviewed.
(b.) Determine the probability that Jude will review Book $V$ immediately after reviewing Book $C$.
(c.) Determine the probability that Jude placed Book $C$ and Book $V$ next to each other.
(d.) Determine the probability that Jude placed Book $C$ first or Book $V$ last.
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
(2.) Question (36.) on Probability 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?
(3.) WASSCE-FM A group consists of 8 boys and 5 girls.
A committee of 7 members is chosen from the group.
Find the probability that the committee is made up of 4 boys and 3 girls.
(4.) Question (30.) on Probability
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?
(5.) Seven students: Students $A, B, C, D, E, F, G$ stand in a linear random order to take a group
photograph.
Each has an equal probability of standing in a certain position provided a linear order is maintained.
(I.) Determine the probability that Student $A$ and Student $C$ stand next to each other.
(II.) Determine the probability that Student $D$ is in the rightmost position.
(III.) Are the two events in (I.) and (II.) independent? Give reasons for your answer.
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:
(6.) ACT At a local post office, on average, $3$ customers are in line when the post office
closes each day.
The probability, $P$, that exactly $n$ customers are in line when the post office closes can be modeled
by the equation:
$P = \dfrac{3^n e^{-3}}{n!}$
Given that $e^{-3} \approx 0.05$, which of the following values is closest to the probability that
exactly $2$ customers
are in line when the post office closes?
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
$
(7.) WASSCE-FM A pool of jurors consists of 1 British, 3 Americans, and 2 Africans.
If 2 jurors are selected one after the other, to sit on jury for a trial at a time, find the probability
that they are:
(a.) 1 American and 1 African
(b.) of different nationalities
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:
(9.) 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?
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
For ACT Students