Question

If a test to detect a disease whose prevalence is 1 out of 1,000 has a false positive rate of 5 percent, what is the chance that a person found to have a positive result actually has the disease?

Answer 1

About 2%

Explanation:

Let call A the case where a person has the disease, and B the case where the person got a positive in the test. The chance that a person found to have a positive result actually has the disease (P(A|B)) can be found using Bayes' theorem:

P(A|B) = P(B|A)*P(A)

P(B)

P(B|A) is the probability of obtaining a positive result in the test in the case that we actually know that we have the disease. This value would be 1. P(A) is the probability of having the disease, and it is said by the problem: 1/1000. P(B) is the probability of obtaining a positive test. Be careful, this positive test can be a real positive test or a false positive test. We just care about obtaing a positive. This would be the probaility of obtaining a real positive plus the probability of obtaining a false positive.

The probability of obtaining a real positive is the probability of having the disease and obtaining a positive. Notice we already established how to calculate this value (P(B|A)*P(A))

The probability of obtaining a false positive is the probability of not having the disease (999/1000) and obtaining a positive (5%).

Then:

P(A|B) = P(B|A)*P(A) = 1* 1/1000 = 0.1962 or about 2%

P(B) 1*1/1000 + 999/1000 * 0.05