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?


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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.