Only one in 1000 adults is afflicted with a rare disease for

Only one in 1,000 adults is afflicted with a rare disease for which a diagnostic test has been developed. This test is such that, when an individual actually has the disease, a positive result will result 99% of the time, while an individual without the disease will show a positive test result only 2% of the time. If a randomly selected individual is tested and the result is positive, what is the probability that the individual has the disease?

Solution

Let

D+, D- = positive or negative to the disease, respectively

T+, T- = postive or negative to the test, respectively

P(D+|T+) = P(D+ n T+)/P(T+)

As

P(D+ n T+) = P(D+) P(T+|D+) = (1/1000)*(0.99) = 0.00099

and

P(T+) = P(D+) P(T+|D+) + P(D-) P(T+|D-) = (1/1000)*(0.99) + (1 - 1/1000)*(0.02) = 0.02097

Thus,

P(D+|T+) = P(D+ n T+)/P(T+) = 0.0472103 [answer]