[tex]\begin{gathered} Let\text{ P(x) represents the probability that }her\text{ family has a ri}sk\text{ of heart attack} \\ P(x)=70\text{\%=}\frac{70}{100}=0.7 \end{gathered}[/tex][tex]\begin{gathered} \text{Let P(y) represent the probability of the reliability of the }stress\text{ test} \\ P(y)=67\text{\%=}\frac{67}{100}=0.67 \end{gathered}[/tex][tex]\begin{gathered} Let\text{ P(x') be the event that her family will NOT have a risk of heart attack} \\ P(x^{\prime})=1-P(x) \\ P(x^{\prime})=1-0.7 \\ P(x^{\prime})=0.3 \end{gathered}[/tex][tex]\text{ity that Erin will not have a heart attack and the test predicts that she will is:}[/tex]
