$\begin{array}{|lcccccccc|}\hline \text{Days}& 0& 1& 2& 3& 4& 5& 6& 7\\ \text{Probability}& 0.68& 0.05& 0.07& 0.08& 0.05& 0.04& 0.01& 0.02\\ \hline\end{array}$

Working out Refer to Exercise 6. Consider the events A = works out at least once and B = works out less than 5 times per week.

(a) What outcomes makeup event A? What is P(A)?

(b) What outcomes make up event B? What is P(B)?

(c) What outcomes make up the event “A and B”? What is P(A and B)? Why is this probability not equal to P(A) · P(B)?

Given probability distribution is

$\begin{array}{|lcccccccc|}\hline \text{Days}& 0& 1& 2& 3& 4& 5& 6& 7\\ \text{Probability}& 0.68& 0.05& 0.07& 0.08& 0.05& 0.04& 0.01& 0.02\\ \hline\end{array}$

Consider $A$as the event that depicts the workout at least once.

The outcomes for the event $A$can be written as:

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

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

localid="1649992396909" $$\begin{gathered} P\left(A\right)=P(X=1)+P(X=2)+\dots ..+P(X=7) \\ =0.05+0.08+\dots ..+0.02 \\ =0.32 \end{gathered}$$

Given probability distribution is

$\begin{array}{|lcccccccc|}\hline \text{Days}& 0& 1& 2& 3& 4& 5& 6& 7\\ \text{Probability}& 0.68& 0.05& 0.07& 0.08& 0.05& 0.04& 0.01& 0.02\\ \hline\end{array}$

Consider $A$as the event that depicts the workout less than $5$times in a week.

The outcomes for the event localid="1649992417188" $B$can be written as:

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

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

localid="1649992420222" $$\begin{gathered} P\left(B\right)=P(X=0)+P(X=1)+\dots ..+P(X=4) \\ =0.68+0.05+\dots +0.05 \\ =0.93 \end{gathered}$$

Given probability distribution is

$\begin{array}{|lcccccccc|}\hline \text{Days}& 0& 1& 2& 3& 4& 5& 6& 7\\ \text{Probability}& 0.68& 0.05& 0.07& 0.08& 0.05& 0.04& 0.01& 0.02\\ \hline\end{array}$

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

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

$P(A$and $B)$can be calculated as:

localid="1649992432843" $$\begin{gathered} P\left(A\text{and}B\right)=P(X=1)+P(X=2)+P(X=3)+P(X=4) \\ =0.05+0.07+0.08+0.05 \\ =0.25 \end{gathered}$$