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An ectopic pregnancy is twice as likely to occur when the pregnant woman is a smoker than when she is a nonsmoker. If 32% of women of childbearing age are smokers, what percentage of women having ectopic pregnancies are smokers?

Attempt: Given the women has had an ectopic pregnancy, we can condition on her being a smoker. So I believe the prob we want in this question is P(woman smoke| she had ectopic preg).

Denote ectopic preg by EP

This is equal to P(EP and woman smoke)/P(EP), which can be rewritten as P(EP|woman smoke)/(P(woman smoke)P(EP)) by Bayes.

Then taking into consideration data in the question, I said the above can further be written as 2P(EP|woman not smoke)/(P(woman smoke)P(EP)). (1)

Using the total Prob law on P(EP) gives P(EP) = P(EP|woman smoke)P(woman smoke) + P(EP|woman did not smoke)P(woman did not smoke), which is equal to P(EP|woman did not smoke)[2P(woman smoke) + P(woman did not smoke)]

Subbing this into (1) and cancelling gives 2/(.32)(0.64 + .68) which is > 1. There is probably a simpler method to mine, but I also want to know where I went wrong. Many thanks.

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Let $E$ denote the event of an ectopic pregnancy and let $S$ denote the event that the woman is a smoker. So far we are given $\Pr(E\mid S) = 2\Pr(E\mid\overline{S}),\ \ \ \Pr(S)=0.32$ We wish to find $P(S\mid E)$. From Bayes' law $\Pr(S\mid E) = \frac{\Pr(E\cap S)}{\Pr(E)}=\frac{\Pr(E\mid S)\Pr(S)}{\Pr(E\mid S)\Pr(S) + \Pr(E\mid \overline{S})\Pr(\overline{S})}$ Making a few substitutions, we have $\frac{\Pr(E\mid S)\Pr(S)}{\Pr(E\mid S)\Pr(S) + 0.5\Pr(E\mid S)\left(1-\Pr(S)\right)}=\frac{2\Pr(S)}{1+\Pr(S)}=\frac{0.64}{1.32}$ The final probability comes down to $\Pr(S\mid E) = \frac{16}{33}\approx 0.485$ Where you went wrong seems to be in applying Bayes' law. You made the substitution $\Pr(S\mid E) = \frac{\Pr(E\cap S)}{\Pr(E)}=\frac{\Pr(E\mid S)}{\Pr(E)\color{red}{\Pr(S)}}$ The term $\Pr(S)$ is on the numeartor, not the denominator. It should be $\Pr(S\mid E) = \frac{\Pr(E\cap S)}{\Pr(E)}=\frac{\Pr(E\mid S)\color{green}{\Pr(S)}}{\Pr(E)}$

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    Ah, so I just made a careless error at the second step. Thank you!2012-11-29