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(A) Let $A$ be the event that the patient has a heart attack. Let $E_{1}$ be the event that the patient chooses the meditation and yoga course,and $E_

Vedclass · Accepted Answer

$\frac{14}{29}$

Vedclass · Answer

(A) Let $A$ be the event that the patient has a heart attack. Let $E_{1}$ be the event that the patient chooses the meditation and yoga course,and $E_{2}$ be the event that the patient chooses the drug prescription. Given $P(E_{1}) = P(E_{2}) = \frac{1}{2}$. The initial probability of a heart attack is $0.40$. If the patient chooses $E_{1}$,the risk is reduced by $30 \%$,so the new probability is $P(A|E_{1}) = 0.40 	imes (1 - 0.30) = 0.40 	imes 0.70 = 0.28$. If the patient chooses $E_{2}$,the risk is reduced by $25 \%$,so the new probability is $P(A|E_{2}) = 0.40 	imes (1 - 0.25) = 0.40 	imes 0.75 = 0.30$. Using Bayes' Theorem,the probability that the patient followed the meditation and yoga course given they had a heart attack is: $P(E_{1}|A) = \frac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1}) + P(E_{2})P(A|E_{2})}$ $P(E_{1}|A) = \frac{\frac{1}{2} 	imes 0.28}{\frac{1}{2} 	imes 0.28 + \frac{1}{2} 	imes 0.30} = \frac{0.28}{0.28 + 0.30} = \frac{0.28}{0.58} = \frac{28}{58} = \frac{14}{29}$.

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Events $E_{1}$ and $E_{2}$ form a partition of the sample space $S$. $A$ is any event such that $P(E_{1}) = P(E_{2}) = \frac{1}{2}$,$P(E_{2} | A) = \frac{1}{2}$,and $P(A | E_{2}) = \frac{2}{3}$. Then $P(E_{1} | A)$ is:

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