Chapter 4.2, Problem 20P

a.

To determine

Find the probability of getting a sum of 7.

Answer to Problem 20P

The probability of getting a sum of 7 is 0.167.

Explanation of Solution

 It is given that two fair dice are rolled, then the sample space is $n\left(s\right)=6^2$.

That is, the following is obtained:

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

The outcomes on the two fair dice are equally likely, mutually exclusive (disjoint), and independent events.

The probability of getting a sum of 7 is as follows:

$\begin{array}{c}P\left(\text{Sum 7}\right)=P\left[\left(1,6\right)or\left(2,5\right)or\left(3,4\right)or\left(4,3\right)or\left(5,2\right)or\left(6,1\right)\right]\\ =P\left(1,6\right)+P\left(2,5\right)+P\left(3,4\right)+P\left(4,3\right)+P\left(5,2\right)+P\left(6,1\right)\\ =\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)\\ =\frac{1+1+1+1+1+1}{36}\\ =\frac{6}{36}\\ =\frac{1}{6}\\ \approx 0.167\end{array}$

Thus, the probability of getting a sum of 7 is 0.167.

b.

To determine

Find the probability of getting a sum of 11.

Answer to Problem 20P

The probability of getting a sum of 11 is 0.055.

Explanation of Solution

The probability of getting a sum of 11 is as follows:

$\begin{array}{c}P\left(\text{Sum 11}\right)=P\left[\left(5,6\right)or\left(6,5\right)\right]\\ =P\left(5,6\right)+P\left(6,5\right)\\ =\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)\\ =\frac{1+1}{36}\\ =\frac{2}{36}\\ =\frac{1}{18}\\ \approx 0.055\end{array}$

Thus, the probability of getting a sum of 11 is 0.055.

c.

To determine

Find the probability of getting a sum of 7 or 11.

Explain whether the outcomes 7 or 11 are mutually exclusive.

Answer to Problem 20P

The probability of getting a sum of 7 or 11 is 0.222.

Explanation of Solution

The outcomes of getting a sum of 7 or 11 are mutually exclusive events because these outcomes cannot be obtained at a time.

From Parts (a) and (b), it is found that $P\left(\text{Sum 7}\right)=\frac{6}{36}$, $P\left(\text{Sum 11}\right)=\frac{2}{36}$.

The probability of getting a sum of 7 or 11 is as follows:

$\begin{array}{c}P\left(\text{Sum 7 }or\text{ 11}\right)=\frac{6}{36}+\frac{2}{36}\\ =\frac{8}{36}\\ =\frac{2}{9}\\ \approx 0.222\end{array}$

Thus, the probability of getting a sum of 7 or 11 is 0.222.