Screening Test, Base Rate and a Second Positive
A district health camp screens 10,000 adults for a disease using a rapid test. The disease prevalence in the camp population is 5%. The test has sensitivity 92% (positive in a diseased person) and specificity 88% (negative in a healthy person). A camp worker reports that 'of those who test positive, 92% will have the disease, because that is the sensitivity.' The medical officer is worried about false positives and wants a proper calculation before sending letters. A second, independent test is available and is planned for anyone who tests positive, with the same sensitivity and specificity. The officer wants to know the probability of disease after one positive test and after two positive tests, and whether the second test is worth the extra cost given the camp's budget.
1. Calculate P(disease | first test positive) and compare it with the worker's 92% claim.
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Use a hypothetical 10,000 people. With 5% prevalence, about 500 have the disease and 9,500 do not. Of the 500 diseased, sensitivity 92% gives 0.92×500=460 true positives. Of the 9,500 healthy, specificity 88% means 12% falsely test positive, giving 0.12×9,500=1,140 false positives. Total positives = 460+1,140=1,600. Therefore P(disease | positive)=460/1,600=0.2875, or 28.75%. The worker's 92% is P(positive | disease), not P(disease | positive). The two conditional probabilities are reversed; the base rate of 5% is crucial. A positive result raises the probability from 5% to about 28.75%, nearly six times the prior, but it is still far from 92%.
2. Calculate P(disease | both tests positive), assuming the tests are independent given disease status.
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Use Bayes' rule or frequencies. With prior P(D)=0.05 and P(H)=0.95. For a diseased person, P(++|D)=0.92×0.92=0.8464. For a healthy person, P(++|H)=0.12×0.12=0.0144. The joint probability of disease and two positives is 0.05×0.8464=0.04232. The joint probability of healthy and two positives is 0.95×0.0144=0.01368. Total probability of two positives = 0.04232+0.01368=0.05600. Hence P(D|++)=0.04232/0.05600≈0.7557, about 75.6%. Equivalently, in 10,000 people, about 500×0.8464≈423 diseased people and 9,500×0.0144≈137 healthy people would produce two positives, giving 423/(423+137)≈0.756. The second positive raises the probability from 28.75% to about 75.6%.
3. The officer says 'a positive result is still wrong more than 70% of the time after one test, so the test is useless.' Evaluate this reasoning.
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The statement confuses 'wrong after one positive' with 'useless'. After one positive, the probability of disease is 28.75%, so the probability of not having disease is 71.25%; that part is numerically right. But the relevant comparison is with the prior probability of 5%. A positive result multiplies the odds of disease by 0.92/0.12≈7.67, raising the probability from 5% to 28.75%. That is a large information gain and is exactly what a screening test is for: it selects a smaller group for confirmatory testing. The second independent test raises the probability to about 75.6%, which is high enough to justify treatment or further investigation. The test is not useless; it is imperfect. The officer should also consider the cost per true case detected and the harm of missing cases, not just the false-positive fraction. A blanket claim that the test is useless ignores the base rate and the purpose of sequential testing.