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Can someone explain how to solve the following stats problem:

68% of students study for an exam. Of those who study, 97% pass. Of those who do not study, 60% pass. What is the probability that a teenager who passes the exam did not study?

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Let $P$ denote the event the student passes, and let $NS$ denote the event the student did not study. We want the conditional probability $\Pr(NS|P)$. By the definition of conditional probability, we have $$\Pr(NS|P)=\frac{\Pr(NS\cap P)}{\Pr(P)}.$$ We want to calculate the two probabilities on the right.

Let's do the hard part first, and find $\Pr(P)$. Passing can happen in two ways: (i) did not study and passed or (ii) studied and passed.

For the probability of (i), the probability a student does not study is $0.32$. Given she does not study, the probability she passes is $0.60$. So the probability of (i) is $(0.32)(0.60)$.

Remark: There is no strong connection between this problem and the binomial distribution.

Similarly, the probability of (ii) is $(0.68)(0.97)$.

For $\Pr(P)$, add the answers to (i) and (ii).

Now we want the numerator, the probability of $NS\cap P$. We have already computed this, it is the probability of (i).

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  • $\begingroup$ This is extremely helpful, thank you. $\endgroup$ – user75133 Apr 30 '13 at 5:02
  • $\begingroup$ You are welcome. Perhaps if you have further questions, you could indicate what you have tried, so that answers can focus on your individual source of difficulty. $\endgroup$ – André Nicolas Apr 30 '13 at 5:05

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