Chapter 6.1, Problem 8E

a.

To determine

To find: The outcomes for the event A and $P\left(A\right)$ .

Answer to Problem 8E

The probability is 0.32.

Explanation of Solution

Given:

The probability distribution is:

Days	0	1	2	3	4	5	6	7
Probability	0.68	0.05	0.07	0.08	0.05	0.04	0.01	0.02

Calculation:

Here. A be the event that depicts the workout at least once.

The outcomes for the event A can be written as:

  $\text{Outcomes = }\left\{1,2,3,4,5,6,7\right\}$

  $P\left(A\right)$ can be calculated as:

  $\begin{array}{c}P\left(A\right)=P\left(X=1\right)+P\left(X=2\right)+\mathrm{.....}+P\left(X=7\right)\\ =0.05+0.08+\mathrm{.....}+0.02\\ =0.32\end{array}$

Thus, the probability is 0.32.

b.

To determine

To find: The outcomes for the event B and $P\left(B\right)$ .

Answer to Problem 8E

The probability is 0.93.

Explanation of Solution

Calculation:

Here. A be the event that depicts the workout less than 5 times in a week.

The outcomes for the event B can be written as:

  $\text{Outcomes = }\left\{0,1,2,3,4\right\}$

  $P\left(B\right)$ can be calculated as:

  $\begin{array}{c}P\left(B\right)=P\left(X=0\right)+P\left(X=1\right)+\mathrm{.....}+P\left(X=4\right)\\ =0.68+0.05+\mathrm{....}+0.05\\ =0.93\end{array}$

Thus, the probability is 0.93.

c.

To determine

To find: The outcomes for the event A and B and compute $P\left(A\text{ and }B\right)$ . Also, explain the reason of $P\left(A\text{ and }B\right)$ not being equal to $P\left(A\right).P\left(B\right)$

Answer to Problem 8E

The probability is 0.25.

Explanation of Solution

Calculation:

Using the information of above parts, the outcomes for the event A and B can be written as:

  $\text{Outcomes = }\left\{1,2,3,4\right\}$

  $P\left(A\text{ and }B\right)$ can be calculated as:

  $\begin{array}{c}P\left(A\text{ and }B\right)=P\left(X=1\right)+P\left(X=2\right)+P\left(X=3\right)+P\left(X=4\right)\\ =0.05+0.07+0.08+0.05\\ =0.25\end{array}$

Here, $P\left(A\text{ and }B\right)\ne P\left(A\right)\times P\left(B\right)$ because the events are not independent.