(C) Let $S$ be the event that a person is a smoker and $NS$ be the event that a person is a non-smoker.Let $D$ be the event that death is due to lung

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(C) Let $S$ be the event that a person is a smoker and $NS$ be the event that a person is a non-smoker.Let $D$ be the event that death is due to lung cancer.Given: $P(S) = 0.20$, $P(NS) = 0.80$, and $P(D) = 0.006$.According to the problem, $P(D|S) = 10 \times P(D|NS)$, which implies $P(D|NS) = \frac{1}{10} P(D|S)$.Using the Law of Total Probability:$P(D) = P(S) \cdot P(D|S) + P(NS) \cdot P(D|NS)$$0.006 = 0.20 \cdot P(D|S) + 0.80 \cdot \left( \frac{1}{10} P(D|S) \right)$$0.006 = 0.20 \cdot P(D|S) + 0.08 \cdot P(D|S)$$0.006 = 0.28 \cdot P(D|S)$$P(D|S) = \frac{0.006}{0.28} = \frac{6}{280} = \frac{3}{140}$.

Class 12 Mathematics Probability Baye's theorem

Medium

$A$ survey of people in a given region showed that $20 \%$ were smokers. The probability of death due to lung cancer given that a person smoked was $10$ times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is $0.006$, what is the probability of death due to lung cancer given that a person is a smoker?

WBJEE 2015 All WBJEE

A
$1 / 140$
B
$1 / 70$
C
$3 / 140$
D
$1 / 10$

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