Chapter 4.2, Problem 19P

a.

To determine

Find the probability of getting a sum of 6.

Answer to Problem 19P

The probability of getting a sum of 6 is 0.139.

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 6 is as follows:

$\begin{array}{c}P\left(\text{Sum }6\right)=P\left[\left(1,5\right)or\left(2,4\right)or\left(3,3\right)or\left(4,2\right)or\left(5,1\right)\right]\\ =P\left(1,5\right)+P\left(2,4\right)+P\left(3,3\right)+P\left(4,2\right)+P\left(5,1\right)\\ =\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)\\ =\frac{1+1+1+1+1}{36}\\ =\frac{5}{36}\\ \approx 0.139\end{array}$

Thus, the probability of getting a sum of 6 is 0.139.

b.

To determine

Find the probability of getting a sum of 4.

Answer to Problem 19P

The probability of getting a sum of 4 is 0.083.

Explanation of Solution

The probability of getting a sum of 4 is as follows:

$\begin{array}{c}P\left(\text{Sum 4}\right)=P\left[\left(1,3\right)or\left(2,2\right)or\left(3,1\right)\right]\\ =P\left(1,3\right)+P\left(2,2\right)+P\left(3,1\right)\\ =\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)+\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)\\ =\frac{1+1+1}{36}\\ =\frac{3}{36}\\ =\frac{1}{12}\\ \approx 0.083\end{array}$

Thus, the probability of getting a sum of 4 is 0.083.

c.

To determine

Find the probability of getting a sum of 6 or 4.

Explain whether the outcomes 6 or 4 are mutually exclusive.

Answer to Problem 19P

The probability of getting a sum of 6 or 4 is 0.222.

Explanation of Solution

It is noticed that the sum of 6 or 4 is mutually exclusive because these outcomes cannot be obtained at the same time.

From Parts (a) and (b), it is found that $P\left(\text{Sum }6\right)=\frac{5}{36}$, $P\left(\text{Sum 4}\right)=\frac{3}{36}$.

The probability of getting a sum of 6 or 4 is as follows:

$\begin{array}{c}P\left(\text{Sum }6or4\right)=\frac{5}{36}+\frac{3}{36}\\ =\frac{8}{36}\\ =\frac{2}{9}\\ \approx 0.222\end{array}$

Thus, the probability of getting a sum of 6 or 4 is 0.222.