Chapter 15.4, Problem 18SRE

a.

To determine

Find the chance that one of the numbers is 2 and the other is 5.

Answer to Problem 18SRE

The chance that one of the numbers is 2 and the other is 5 is 0.067.

Explanation of Solution

Calculation:

From the information, the sample space is as follows:

$\begin{array}{c}S=\left\{\begin{array}{l}\left(1,1\right)\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right)\\ \left(2,1\right)\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right)\\ \left(3,1\right)\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right)\\ \left(4,1\right)\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right)\\ \left(5,1\right)\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right)\end{array}\right\}\\ N\left(S\right)=30\end{array}$

The chance that one of the numbers is 2 and the other is 5 is as follows:

$\begin{array}{c}N\left(\begin{array}{l}\text{one of the numbers is 2 }\\ \text{and the other is 5 }\end{array}\right)=\left\{\left(2,5\right),\left(5,2\right)\right\}\\ =2\\ P\left(\begin{array}{l}\text{one of the numbers is 2 }\\ \text{and the other is 5 }\end{array}\right)=\frac{N\left(\begin{array}{l}\text{one of the numbers is 2 }\\ \text{and the other is 5 }\end{array}\right)}{N\left(S\right)}\\ =\frac{2}{30}\\ =0.067\end{array}$

Thus, the chance that one of the numbers is 2 and the other is 5 is 0.067.

b.

To determine

Find the chance that the sum of the numbers is 7.

Answer to Problem 18SRE

The chance that the sum of the numbers is 7 is 0.167.

Explanation of Solution

Calculation:

The chance that the sum of the numbers is 7 is as follows:

$\begin{array}{c}N\left(\text{Sum of thenumbers is 7}\right)=\left\{\left(1,6\right),\left(2,5\right),\left(3,4\right),\left(4,3\right),\left(5,2\right)\right\}\\ =5\\ P\left(\text{Sum of thenumbers is 7}\right)=\frac{N\left(\text{Sum of thenumbers is 7}\right)}{N\left(S\right)}\\ =\frac{5}{30}\\ =0.167\end{array}$

Thus, the chance that the sum of the numbers is 7 is 0.167.

c.

To determine

Find the chance that one number is bigger than twice the other.

Answer to Problem 18SRE

The chance that one number is bigger than twice the other is 0.367.

Explanation of Solution

Calculation:

The chance that one number is bigger than twice the other is as follows:

$\begin{array}{c}N\left(\begin{array}{l}\text{one number is bigger }\\ \text{than twice the other}\end{array}\right)=\left\{\begin{array}{l}\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,5\right),\left(2,6\right),\left(3,1\right),\left(4,1\right),\left(4,2\right)\\ \left(5,1\right),\left(5,2\right)\end{array}\right\}\\ =11\\ P\left(\begin{array}{l}\text{one number is bigger }\\ \text{than twice the other}\end{array}\right)=\frac{N\left(\begin{array}{l}\text{one number is bigger }\\ \text{than twice the other}\end{array}\right)}{N\left(S\right)}\\ =\frac{11}{30}\\ =0.367\end{array}$

Thus, the chance that one number is bigger than twice the other is 0.367.