Consider a population that contains 95 nonusers of a control

Consider a population that contains 95% non-users of a controlled-substance(i.e. drugs).To test for the substance, a test that is 98% accurate when the test patient is actually using drugs is used. This same test is known to yield a positive response when the test patient is not using drugs 1% of the time. Find the probability of a patient being a non-user if he or she tested positive for drugs

Solution

Let

T+ = test positive
T- = test negative
D+ = used drugs
D- = did not use drugs

Thus,

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

As

P(D- n T+) = P(D-) P(T+|D-) = 0.95*0.01 = 0.0095

P(T+) = P(D-) P(T+|D-) + P(D+) P(T+|D+) = 0.95*0.01 + (1-0.95)*0.98 = 0.0585

Thus,

P(D-|T+) = P(D- n T+) / P(T+) = 0.0095 / 0.0585 = 0.162393162 [ANSWER]