Example 2.31 Incidence of a rare disease. Only 1 in 1000 adults is afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease, a positive result will occur 99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time (the sensitivity of this test is 99% and the specificity is 98%; in conrast, the Sept. 22, 2012 issue of The Lancet reports that the first at-home HIV test has a sensitivity of only 92% and a specificity of 99.98%). If a randomly selected individual is tested and the result is positive, what is the probability that the individual has the disease? To use Bayes' theorem, let A, = individual has the disease, A, = individual does not have the disease, and B = positive test result. Then P(A,) = P(A,) = P(B|A,) = , and P(B|A,) = The tree diagram for this problem is below. P(A, N B) = .00099 99 B= +Test 01 001 B'= -Test A, = Has disease - P(A2 0 B) = 01998 .999 A, = Doesn't have disease 02 B= +Test 98 B'= -Test Tree diagram for the rare-disease problem Next to each branch corresponding to a positive test result, the --Select-- rule yields the recorded probabilities. Therefore, P(B) = 0.00099 + 0.01998 = (entered to five decimal places), from which we have P(A, IB) = PA, n B) 0.00099 P(B) |(rounded to three decimal places) 0.02097 This result seems counterintuitive; the diagnostic test appears so accurate we expect someone with a positive test result to be highly likely to have the disease, whereas the computed conditional probability is only 0.047. However, the rarity of the disease implies that most positive test results arise from errors rather than from diseased individuals. The probability of having the disease has --Select-. v by a multiplicative factor of 47 (from prior test with much smaller error rates is needed. to posterior 0.047); but to get a further increase in the posterior probability, a diagnostic