Consider the probabilities of people taking HIV tests. Assume that the true probability of having HIV for all people who take the test is 4%. If HIV tests give a positive result in 99% of the cases in which the person has HIV and give a negative result in 99% of the cases in which the person does not have HIV, what is the probability that a person who gets a positive test result does not have HIV?

Question

Quizplus · Accepted Answer

This is a classic example of Bayes' theorem. Let's denote having HIV as H and testing positive as +. We're looking for P(H'|+), the probability of not having HIV given a positive test result. Given: P(H) = 0.04, P(+|H) = 0.99, P(-|H') = 0.99 (which implies P(+|H') = 0.01, since if you don't have HIV, there's a 1% chance you'll still test positive). Using Bayes' theorem: P(H'|+) = P(+|H')P(H') / [P(+|H)P(H) + P(+|H')P(H')]. Plugging in the numbers: P(H'|+) = (0.01 * 0.96) / [(0.99 * 0.04) + (0.01 * 0.96)] = 0.0096 / (0.0396 + 0.0096) = 0.0096 / 0.0492 ≈ 0.195.

Consider the Probabilities of People Taking HIV Tests