A medical test has been designed to detect the presence of a certain disease. Among those who have the disease, the probability that the disease will be detected by the test is 0.97. However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is 0.05. Suppose 20% of the people who were referred to a clinic for the test did in fact have the disease. If the test administered to an individual from this group is positive, what is the probability that the person actually has the disease?
A) 0.471
B) 0.545
C) 0.829
D) 0.857

Question

Quizplus · Accepted Answer

This is a classic example of applying Bayes' theorem. Let's denote having the disease as D and testing positive as T. We're given P(T|D) = 0.97, P(T|¬D) = 0.05, and P(D) = 0.20. We want to find P(D|T), which is given by Bayes' theorem as P(D|T) = [P(T|D) * P(D)] / [P(T|D) * P(D) + P(T|¬D) * P(¬D)]. Substituting the given values, we get P(D|T) = (0.97 * 0.20) / (0.97 * 0.20 + 0.05 * 0.80) = 0.194 / (0.194 + 0.04) = 0.194 / 0.234 ≈ 0.829.

A Medical Test Has Been Designed to Detect the Presence