25% of the population are smokers. A smoker has 27 times more chances to develop lung cancer than a non-smoker. A person is diagnosed with lung cancer and the probability that this person is a smoker is $\frac{k}{10}$. Then the value of k is ___.
Step-by-Step Solution
Key Concept: Apply Bayes' theorem with $P(E_1) = 1/4$ (smoker), $P(E_2) = 3/4$ (non-smoker), $P(\text{cancer}|E_1) = 27 \times P(\text{cancer}|E_2)$
$P(E_1) = 1/4$, $P(E_2) = 3/4$, $P(C|E_1)/P(C|E_2) = 27$. Let $P(C|E_2) = p$, $P(C|E_1) = 27p$. $P(E_1|C) = \frac{(1/4)(27p)}{(1/4)(27p)+(3/4)p} = \frac{27/4}{27/4+3/4} = \frac{27}{30} = \frac{9}{10}$. So $k=9$. Answer: 9
Correct Answer: 9