Chapter 11.8, Problem 27E

a.

To determine

TO-DETERMINE the probability of getting a $6$ when two number cubes are rolled.

Answer to Problem 27E

The probability of getting at least one $6$ is $\text{P}\left(\text{A}\right)=\frac{1}{12}$ .

Explanation of Solution

Given:

Consider, a red number cube and a blue number cube is rolled and A is the event of at least one number cube showing a six.

Calculation:

The total number of possible outcomes when two number cubes are rolled is 36.

The probability of getting at least one $6$ is $\text{P}\left(\text{6 on red cube, no 6 on blue cube}\right)$ , $\text{P}\left(\text{no 6 on red cube, 6 on blue cube}\right)$ , $\text{P}\left(\text{6 on both cubes}\right)$ .

The number of possible outcomes of getting at least one $6$ is $3$

  $\begin{array}{l}\text{P}\left(\text{A}\right)=\frac{3}{36}\\ \text{P}\left(\text{A}\right)=\frac{1}{12}\left[\text{simplify}\right]\end{array}$

Hence, the probability of getting at least one $6$ is $\text{P}\left(\text{A}\right)=\frac{1}{12}$ .

ii.

To determine

TO-DETERMINE the probability of getting a $6$ when two number cubes are rolled.

Answer to Problem 27E

Hence, the probability of getting at least one $6$ is $\text{P}\left(\text{A}\right)=\frac{1}{12}$

Explanation of Solution

Given:

Consider, a red number cube and a blue number cube is rolled and A is the event of at least one number cube showing a six.

Calculation:

The total number of possible outcomes when two number cubes are rolled is 36.

  $\text{P}\left(\text{not A}\right)$ is the events of not getting a $6$

The number of outcomes of not getting a $6$ is $33$

  $\begin{array}{l}\text{P}\left(\text{not A}\right)=\frac{33}{36}\\ \text{P}\left(\text{not A}\right)=\frac{11}{12}\left[\text{simplify}\right]\end{array}$

The probability of not getting a $6$ is $\text{P}\left(\text{not A}\right)=\frac{11}{12}$ .

The probability of getting a $6$ can be calculated by

  $\begin{array}{l}\text{P}\left(\text{A}\right)=1-\text{P}\left(\text{not A}\right)\left[\text{probability of complementary events}\right]\\ \text{P}\left(\text{A}\right)=1-\frac{11}{12}\left[\text{substitute}\right]\\ \text{P}\left(\text{A}\right)=\frac{1}{12}\end{array}$

Hence, the probability of getting at least one $6$ is $\text{P}\left(\text{A}\right)=\frac{1}{12}$ .