Bayes' Theorem

Bayes' theorem combines a prior prevalence with test likelihoods to give the posterior probability of disease given a positive result, through P(D | +) = P(+ | D)P(D) / [P(+ | D)P(D) + P(+ | D')P(D')]. This screening problem is a celebrated JEE Advanced application. The prior is P(D) = 0.01 and P(D') = 0.99; the sensitivity gives P(+ | D) = 0.99 and the false positive rate gives P(+ | D') = 0.02. The numerator is 0.99 × 0.01 = 0.0099, and the denominator adds the false-positive contribution 0.02 × 0.99 = 0.0198, totalling 0.0099 + 0.0198 = 0.0297. Thus P(D | +) = 0.0099 / 0.0297 = 99/297, which reduces to 1/3. Option 99/100 mistakes the answer for the sensitivity. Option 1/2 assumes equal priors, ignoring the low prevalence. Option 99/101 mishandles the false-positive term. The theorem shows how rare diseases produce many false positives, because the healthy group, though individually unlikely to test positive, is so much larger that its few false alarms outnumber the true detections. Concretely, among a population of 10000 people, about 99 of the 100 diseased test positive while about 198 of the 9900 healthy also test positive, so positives are dominated by the healthy. Plausibility check: despite a 99% sensitive test, the posterior 99/297 ≈ 0.33 stays low, which is the expected counterintuitive base-rate effect that arises whenever the disease prevalence is much smaller than the false-positive rate is large.