A positive test for a rare disease

Question

A disease affects 1 in 1{,}000 people. A test detects it 99\% of the time when present (sensitivity 0.99) and gives a false positive 5\% of the time when absent. You test positive. What is the probability you actually have the disease?

Accepted Answer

Apply Bayes in odds form. Prior odds of disease are 1 : 999. The likelihood ratio of a positive result is \frac{P(+ \mid 	ext{disease})}{P(+ \mid 	ext{healthy})} = \frac{0.99}{0.05} = 19.8. Posterior odds = 1 : 999 	imes 19.8 = 19.8 : 999, so P(	ext{disease} \mid +) = \frac{19.8}{19.8 + 999} = \frac{19.8}{1018.8} \approx 0.0194 = 1.9\%. The direct form agrees: \frac{0.001 	imes 0.99}{0.001 	imes 0.99 + 0.999 	imes 0.05} = \frac{0.00099}{0.05094} \approx 0.019. Despite a "99\% accurate" test, you are still only about 2\% likely to be sick, this is base-rate neglect: the disease is so ra…

A positive test for a rare disease

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