Solution
We are given
Probability of eating out twice in a week = 15% = 0.15
Probability of eating out once in a week = 35% = 0.35
Probability of not eating out in a week = 50% = 0.50
Let X be a random variable of the number of times Piper eats out in a week
So we have the table
Note: The Formula For finding the Expected value E(X) is given by
[tex]E(X)=\sum ^{}_{}xp(x)[/tex]Substituting we get
[tex]\begin{gathered} E(X)=0(0.50)+1(0.35)+2(0.15) \\ E(X)=0+0.35+0.30 \\ E(X)=0.65 \end{gathered}[/tex]Therefore, the expected value is
[tex]E(X)=0.65[/tex]