Chapter 15.10, Problem 5WE

(a)

To determine

To find: The probability of $A$ , $B$ , $A\cup B$ and $A\cap B$ .

Answer to Problem 5WE

The probability of $P\left(A\right)=\frac{1}{6}$ , $P\left(B\right)=\frac{5}{36}$ , $P\left(A\cap B\right)=\frac{1}{36}$ and $P\left(A\cup B\right)=\frac{5}{18}$ .

Explanation of Solution

Given information:

The given information is the event $A$ that the red die shows $2$

The event $B$ that the sum of the numbers showing is $8$

Calculation:

The favorable outcomes are $\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right)$ and $\left(2,6\right)$ . Thus there are $6$ favorable outcomes out of $36$ possible outcomes.

  $\begin{array}{l}P\left(A\right)=\frac{6}{36}\\ P\left(A\right)=\frac{1}{6}\end{array}$

The Probability of the event $A$ that the red die shows is $\frac{1}{6}$ .

The event $B$ occurs when the sum of the numbers on the two dice is $8$ . The sample space of such an event will be $\left(2,6\right),\left(3,5\right),\left(4,4\right),\left(5,3\right),\left(6,2\right)$ . Thus, there are $5$ favorable outcomes out of $36$ possible outcomes

The event $B$ that the sum of the numbers showing is $\left(2,6\right),\left(3,5\right),\left(4,4\right),\left(5,3\right),\left(6,2\right)$ .

The Probability of the event $B$ that the sum of the numbers showing is $\frac{5}{36}$ .

  $P\left(B\right)=\frac{5}{36}$

Now, find $P\left(A\cap B\right)$ .

In this case, $A\cap B$ are the event that the red die shows $2$ and the sum of the numbers on the two dice shows $8$ .

The sample space of $A\cap B$ is $\left(2,6\right)$ Thus,

  $P\left(A\cap B\right)=\frac{1}{36}$

Find $P\left(A\cup B\right)$ .

  $P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

Now,

  $\begin{array}{l}P\left(A\cup B\right)=\frac{1}{6}+\frac{5}{36}-\frac{1}{36}\\ P\left(A\cup B\right)=\frac{6+5-1}{36}\\ P\left(A\cup B\right)=\frac{10}{36}\end{array}$

  $P\left(A\cup B\right)=\frac{5}{18}$

Thus, the probability of $P\left(A\right)=\frac{1}{6}$ , $P\left(B\right)=\frac{5}{36}$ , $P\left(A\cap B\right)=\frac{1}{36}$ and $P\left(A\cup B\right)=\frac{5}{18}$ .

(b)

To determine

To find: The $A$ and $B$ are independent events.

Answer to Problem 5WE

The events $A$ and $B$ are not independent.

Explanation of Solution

Given information:

The given information is $P\left(A\right)=\frac{1}{6}$ , $P\left(B\right)=\frac{5}{36}$ , $P\left(A\cap B\right)=\frac{1}{36}$ and $P\left(A\cup B\right)=\frac{5}{18}$ .

Calculation:

Independent Event: When the outcome of one event doesn’t affect the outcome of others.

  $P\left(A\cap B\right)=P\left(A\right)\times P\left(B\right)$

  $\begin{array}{l}P\left(A\cap B\right)=P\left(A\right)\times P\left(B\right)\\ \frac{1}{36}\ne \frac{1}{6}\times \frac{5}{36}\\ \frac{1}{36}\ne \frac{5}{216}\end{array}$

Since, $P\left(A\cap B\right)\ne P\left(A\right)\times P\left(B\right)$ the events $A$ and $B$ are not independent

Therefore, the event $A$ and $B$ are not independent.