Testing positive twice in a row

Question

A disease affects 1\% of a population. A test catches 90\% of true cases (sensitivity) and clears 90\% of healthy people (specificity). A person takes the test twice, the two results are independent given their true status, and both come back positive. What is the probability that they have the disease?

Accepted Answer

Independent evidence multiplies in odds form. Start with prior odds of disease = 1{:}99. Each positive test multiplies the odds by the likelihood ratio 	ext{LR} = \frac{	ext{sensitivity}}{1 - 	ext{specificity}} = \frac{0.90}{0.10} = 9. Two independent positives multiply by 9 twice: 	ext{posterior odds} = 1{:}99 	imes 9 	imes 9 = 81{:}99. Converting to a probability: P(	ext{disease} \mid ++) = \frac{81}{81 + 99} = \frac{81}{180} = 0.45 = 45\%. As a check, update one step at a time. After the first positive, odds 1{:}99 	imes 9 = 9{:}99 = 1{:}11, so about 8.3\%. Feed that in as the new …

Testing positive twice in a row

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