# Probability with percentages

An ectopic pregnancy is twice as likely to develop for a woman who is a smoker as opposed to nonsmoker. If 28% of women of childbearing age are smokers what percentage of ectopic pregnancies are smokers?

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Hint: Work out $P(E)$ using the law of total probability $$P(E) = P(E\mid S)P(S)+P(E\mid S^c)P(S^c)$$ and the given fact that $P(E\mid S) = 2P(E \mid S^c)$. The answer will be a function of $P(E\mid S)$. Then use Bayes' formula $$P(S\mid E) = \frac{P(E\mid S)P(S)}{P(E\mid S)P(S)+P(E\mid S^c)P(S^c)}$$ and you will see that the unknown vaue of $P(E\mid S)$ will factor right out. – Dilip Sarwate Sep 29 '12 at 2:58

We assume that smokers are just as likely to get pregnant as non-smokers. This assumption is not necessarily reasonable, they may be too busy smoking to get pregnant. But without this kind of assumption we cannot solve the problem.

Let $E$ be the event "ectopic pregnancy," let $S$ be the event "smoker," and let $NS$ be the event "non-smoker." We have been asked to find the conditional probability $\Pr(S|E)$. By the usual formula we have $$\Pr(S|E)=\frac{\Pr(S\cap E}{\Pr(E)}.$$ Now we need to compute the probabilities on the right-hand side. Let $\Pr(E|NS)=p$. Then $\Pr(E|S)=2p$.

The event $E$ can happen in two disjoint ways: (i) the person is a smoker and has ectopic pregnancy or (ii) the person is a non-smoker and has an ectopic pregnancy.

The probability of (i) is $(0.28)(2p)$, and the probability of (ii) is $(0.72)(p)$. So $$\Pr(E)=(0.28)(2p)+(0.72)(p).\tag{1}$$ Easily, $\Pr(S\cap E)=(0.28)(2p)$.

Do the division asked for by Formula $(1)$. The $p$'s cancel.

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Thank u for walking thru. Very helpful. – kimj Sep 29 '12 at 4:50
@kimj: You pressed the spacebar before finishing your message. I get $43.75\%$. – André Nicolas Sep 29 '12 at 4:54