Probability
Bayes Theorem
MMTS_Full_Test_09
Grade 12
Question:
There are two townships $A$ and $B$ in a city containing 40\% and 60\% of the population respectively. 15\% of the total population suffer from heart disease. $P(\text{heart disease}|A) = 6P(\text{heart disease}|B)$. A person randomly diagnosed turns out to be free from heart disease; then the probability that he lives in township $B$ is
$\frac{28}{85}$
$\frac{38}{85}$
$\frac{57}{85}$
$\frac{47}{85}$
Step-by-Step Solution
Key Concept: Let $P(H|B)=p$; $P(H|A)=6p$. Total: $0.4\cdot 6p+0.6\cdot p=0.15\Rightarrow p=1/18$. Use Bayes for $P(B|\bar{H})$.
$0.4(6p)+0.6p=0.15\Rightarrow 2.4p+0.6p=0.15\Rightarrow p=0.05$. $P(H|A)=0.3$, $P(H|B)=0.05$. $P(\bar{H}|A)=0.7$, $P(\bar{H}|B)=0.95$. $P(\bar{H})=0.4\times 0.7+0.6\times 0.95=0.28+0.57=0.85$. $P(B|\bar{H})=\frac{0.6\times 0.95}{0.85}=\frac{0.57}{0.85}=\frac{57}{85}$.
Correct Answer: C