Exercise 2.3 (Smoking and cancer)

Question

According to the CDC (Centers for Disease Control and Prevention), men who smoke are 23 times more likely to develop lung cancer than men who don’t smoke. Also according to the CDC, 21.6\% of men in the U.S. smoke. What is the probability that a man in the U.S. is a smoker, given that he develops lung cancer?

fifthist x · Accepted Answer

Let $S$ be event that a man in the US is a smoker and $C$ be event man has
cancer.
From problem conditions:
\begin{align*}
	&P(S) = 0.216\
	&P(C|S) = 23P(C|S^c)\
	&P(C|S^c) = \frac{1}{23} P(C|S)\
\end{align*}
Lets use Bayes' theorem
\begin{align*}
	P(S|C) & = \frac{P(S)P(C|S)}{P(C)}\
	       & = \frac{P(S)P(C|S)}{P(S)P(C|S) + P(S^c)P(C|S^c)}\
	       & = \frac{P(S) \cdot 23P(C|S^c)}{P(S)\cdot 23P(C|S^c) + P(S^c)P(C|S^c)}\
	       & = \frac{P(S)}{23P(S) + P(S^c)}\
	       & = \frac{23 \cdot 0.216}{23 \cdot 0.216 + 0.784}\
	       & \approx 0.864
\end{align*}

Exercise 2.3 (Smoking and cancer)

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Exercise 2.3 (Smoking and cancer)

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