Chapter 6.3, Problem 26E

a.

To determine

List the 23 possible outcomes in the sample space other than $\left(2,4,3,1\right)$ output.

Answer to Problem 26E

The 23 possible outcomes in the sample space are given as follows:

$\begin{array}{l}\left(1,2,3,4\right),\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,3,4,2\right),\left(1,4,2,3\right),\left(1,4,3,2\right),\\ \left(2,1,3,4\right),\left(2,1,4,3\right),\left(2,3,1,4\right),\left(2,3,4,1\right),\left(2,4,1,3\right),\left(3,1,2,4\right),\\ \left(3,1,4,2\right),\left(3,2,1,4\right),\left(3,2,4,1\right),\left(3,4,1,2\right),\left(3,4,2,1\right),\left(4,1,2,3\right),\\ \left(4,1,3,2\right),\left(4,2,3,1\right),\left(4,2,3,1\right),\left(4,3,1,2\right),\left(4,3,2,1\right)\end{array}$.

Explanation of Solution

Calculation:

The given information is that there are four copies of books that are left under the desk. The professor distributes to four students who lost their books.

Sample Space:

A sample space is the set of all possible outcomes of a random experiment.

There are four copies of books that are left under the desk. The first, second, third, and fourth numbers in the sample point represent the book that are received by students 1, 2, 3, and 4 respectively. For example, outcome $\left(2,3,1,4\right)$ represent first student got second book, second one got third book, third one got first book, and the fourth one got fourth book. The sample space of the experiment is given below:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,2,3,4\right),\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,3,4,2\right),\left(1,4,2,3\right),\left(1,4,3,2\right),\\ \left(2,1,3,4\right),\left(2,1,4,3\right),\left(2,3,1,4\right),\left(2,3,4,1\right),\left(2,4,1,3\right),\left(2,4,3,1\right),\\ \left(3,1,2,4\right),\left(3,1,4,2\right),\left(3,2,1,4\right),\left(3,2,4,1\right),\left(3,4,1,2\right),\left(3,4,2,1\right),\\ \left(4,1,2,3\right),\left(4,1,3,2\right),\left(4,2,3,1\right),\left(4,2,3,1\right),\left(4,3,1,2\right),\left(4,3,2,1\right)\end{array}\right\}$

Therefore, 23 possible outcomes in the sample space are given as follows:

$\begin{array}{l}\left(1,2,3,4\right),\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,3,4,2\right),\left(1,4,2,3\right),\left(1,4,3,2\right),\\ \left(2,1,3,4\right),\left(2,1,4,3\right),\left(2,3,1,4\right),\left(2,3,4,1\right),\left(2,4,1,3\right),\left(3,1,2,4\right),\\ \left(3,1,4,2\right),\left(3,2,1,4\right),\left(3,2,4,1\right),\left(3,4,1,2\right),\left(3,4,2,1\right),\left(4,1,2,3\right),\\ \left(4,1,3,2\right),\left(4,2,3,1\right),\left(4,2,3,1\right),\left(4,3,1,2\right),\left(4,3,2,1\right)\end{array}$.

b.

To determine

List the outcomes in the event ‘exactly two of the books are returned to their correct owners’.

Compute the probability of the event ‘exactly two of the books are returned to their correct owners’.

Answer to Problem 26E

The outcomes in the event ‘exactly two of the books are returned to their correct owners’ is $\left\{\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,4,3,2\right),\left(2,1,3,4\right),\left(3,2,1,4\right),\left(4,2,3,1\right)\right\}$.

The probability of the event is $\frac{1}{4}$ .

Explanation of Solution

Calculation:

The probability of any Event A is given below:

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

Event A denotes that exactly two students receive their own books. $\left(1,2,3,4\right)$. It is found that there are 6 outcomes with exactly two students who receive their own books and are $\left\{\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,4,3,2\right),\left(2,1,3,4\right),\left(3,2,1,4\right),\left(4,2,3,1\right)\right\}$.

Therefore, the outcomes those are in Event A are given as follows:

$A=\left\{\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,4,3,2\right),\left(2,1,3,4\right),\left(3,2,1,4\right),\left(4,2,3,1\right)\right\}$

Among 24 outcomes in the sample space, 6 are in Event A. Substitute these values in the above equation.

The probability that exactly two students receive their own books is calculated as follows:

$\begin{array}{c}P\left(A\right)=\frac{6}{24}\\ =\frac{1}{4}\end{array}$

Thus, the probability that exactly two students receive their own books is $\frac{1}{4}$.

c.

To determine

Compute the probability that exactly one student receives his or her own book.

Answer to Problem 26E

The probability that exactly one student receives his or her own book is $\frac{1}{3}$ .

Explanation of Solution

Calculation:

It is given that there are four copies of the books that are left under the desk. The professor distributes randomly to four students who lost their books.

The probability of any Event A is given below:

$P\left(A\right)=\frac{\text{Number of outcomes in }A}{\text{Total number of outcomes in the sample space}}$

From Exercise 34, it is clear that the sample space of the experiment is given below:

$\text{Sample space}=\left\{\begin{array}{l}\left(1,2,3,4\right),\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,3,4,2\right),\left(1,4,2,3\right),\left(1,4,3,2\right),\\ \left(2,1,3,4\right),\left(2,1,4,3\right),\left(2,3,1,4\right),\left(2,3,4,1\right),\left(2,4,1,3\right),\left(2,4,3,1\right),\\ \left(3,1,2,4\right),\left(3,1,4,2\right),\left(3,2,1,4\right),\left(3,2,4,1\right),\left(3,4,1,2\right),\left(3,4,2,1\right),\\ \left(4,1,2,3\right),\left(4,1,3,2\right),\left(4,2,3,1\right),\left(4,2,3,1\right),\left(4,3,1,2\right),\left(4,3,2,1\right)\end{array}\right\}$

Event A denotes that exactly one student receives his or her own book. That is, either the first, second, third or fourth student receives his or her own book but exactly one of the four receives their own book.

$A=\left\{\begin{array}{l}\left(1,3,4,2\right),\left(1,4,2,3\right),\left(3,2,4,1\right),\left(4,2,1,3\right),\\ \left(2,4,3,1\right),\left(4,1,3,2\right),\left(2,3,1,4\right),\left(3,1,2,4\right)\end{array}\right\}$

Among 24 outcomes in the sample space, 8 are in Event A. Substitute these values in the above equation.

The probability that exactly one student receives his or her own book is calculated as follows:

$\begin{array}{c}P\left(A\right)=\frac{8}{24}\\ =\frac{1}{3}\end{array}$

Thus, the probability that exactly one student receives his or her own book is $\frac{1}{3}$ .

d.

To determine

Find the probability that exactly three of the four students receive their own books.

Answer to Problem 26E

The probability that exactly three of the four students receive their own books is 0.

Explanation of Solution

Calculation:

Event B denotes that exactly three students receive their own books.

If three of the students got their own books then automatically 4^th student will also get her or his own book.

Hence, it is not possible that exactly three students receive their own books. Among 24 outcomes in the sample space, none of the outcomes are in Event B. Substitute these values in the above equation.

The probability that exactly three of the four students receive their own books is calculated as follows:

$\begin{array}{c}P\left(B\right)=\frac{0}{24}\\ =0\end{array}$

Thus, the probability that exactly three of the four students receive their own books is 0.

e.

To determine

Compute the probability that at least two students receive their own books.

Answer to Problem 26E

The probability that at least two students receive their own books is $\frac{7}{24}$.

Explanation of Solution

Calculation:

The Addition Rule for Mutually Exclusive Events:

If Events A, B, and C are three mutually exclusive events, then the addition rule is stated as follows:

$P\left(A\text{or}B\text{or}C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)$

The required probability is given below:

$P\left(\begin{array}{l}\text{at least two receive}\\ \text{ their own books}\end{array}\right)=\left[\begin{array}{l}P\left(\begin{array}{l}\text{exactly two}\text{receive}\\ \text{ their own books}\end{array}\right)+P\left(\begin{array}{l}\text{exactly three}\text{receive }\\ \text{their own books}\end{array}\right)+\\ P\left(\begin{array}{l}\text{exactly four}\text{receive }\\ \text{their own books}\end{array}\right)\end{array}\right]$

From Part (b), it is clear that the probability that exactly three students receive their own books is 0. There is only one possibility that all four receives their own books is $\left(1,2,3,4\right)$. It is found that there are 6 outcomes with exactly two students who receive their own books and those are given below: $\left\{\left(1,2,4,3\right),\left(1,3,2,4\right),\left(1,4,3,2\right),\left(2,1,3,4\right),\left(3,2,1,4\right),\left(4,2,3,1\right)\right\}$. Substitute these values in the above equation.

Therefore, the required probability is given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{at least two receive}\\ \text{ their own books}\end{array}\right)=\frac{6}{24}+0+\frac{1}{24}\\ =\frac{7}{27}\end{array}$.

Thus, the probability that at least two students receive their own books is $\frac{7}{24}$.