Consider the probabilities of people taking HIV tests. Assume that the true probability of having HIV for all people who take the test is 3%. 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 truly has HIV?

Question

Quizplus · Accepted Answer

This is a classic example of applying Bayes' theorem. The probability that a person who tests positive actually has HIV is calculated as follows: P(HIV|positive) = [P(positive|HIV) * P(HIV)] / [P(positive|HIV) * P(HIV) + P(positive|no HIV) * P(no HIV)]. Plugging in the numbers: [0.99 * 0.03] / [0.99 * 0.03 + 0.01 * 0.97] = 0.747, which rounds to approximately 0.75, making option B the correct answer.

Consider the Probabilities of People Taking HIV Tests