A positive test for a rare disease

Question

A disease affects 1\% of a population. A test for it is 99\% accurate in both directions: it is positive for 99\% of sick people (sensitivity) and negative for 99\% of healthy people (specificity). A randomly selected person tests positive. What is the probability that they actually have the disease?

Accepted Answer

Use natural frequencies on a population of 10{,}000: - Sick: 1\% 	imes 10{,}000 = 100 people; of these, 99\% 	imes 100 = 99 test positive (true positives). - Healthy: 9{,}900 people; of these, 1\% 	imes 9{,}900 = 99 test positive (false positives). Among the 99 + 99 = 198 positives, exactly 99 are sick: P(	ext{disease} \mid +) = \frac{99}{198} = \frac{1}{2}. Formally, by Bayes' theorem: P(D \mid +) = \frac{P(+ \mid D)P(D)}{P(+ \mid D)P(D) + P(+ \mid D^c)P(D^c)} = \frac{0.99 	imes 0.01}{0.99 	imes 0.01 + 0.01 	imes 0.99} = \frac{1}{2}. Despite the "99% accurate" test, a positive result …

A positive test for a rare disease

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