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.