91影视

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

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

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