Question

A company administers a drug test to its job applicants as a condition of employment; if a person fails the drug test the company will not hire them.

Suppose the drug test is 77% sensitive and 75% specific. That is, the test will produce 77% true positive results for drug users and 75% true negative results for non-drug users. Suppose that 9% of potential hires are use drugs. If a randomly selected job applicant tests positive, what is the probability he or she is a user? Make sure that your answer is between 0 and 1.

Answer 1

As we are given here that:  drug test is 77% sensitive and 75% specific, therefore we have here:
P( + | drugs taken) = 0.77
P( - | drugs not taken) = 0.75

Also, we are given here that:
P( drugs taken) = 0.09, therefore P(drugs not taken) = 1 - 0.09 = 0.91

Using law of total probability, we have here:
P(+) = P( + | drugs taken) P( drugs taken) + P( + | drugs not taken)P(drugs not taken)
P(+) = 0.77*0.09 + (1 - 0.75)*0.91 = 0.2968

Using Bayes theorem, the probability that he or she is a user given that the test was positive is computed as:

P( drugs taken | +) = P( + | drugs taken)P( drugs taken)/ P(+) = 0.77*0.09 / 0.2968 = 0.2335

Therefore 0.2335 is the required conditional probability here.