Final answer:
Using Bayes' theorem, we calculate the probability of an employee being a drug user given a positive test result, with a sensitivity of 98% and specificity of 91%, and drug use prevalence of 7%. The result is approximately 44.19%, indicating a lower than expected certainty for positive test results.
Step-by-step explanation:
Using Bayes' theorem, we can calculate the probability that an employee who tested positive is actually a drug user. To do this, we'll denote U as the event that an employee uses drugs and T+ as the event that an employee tests positive for drugs. We're given that the sensitivity of the test (P(T+|U)) is 98%, the specificity (P(T-|U')) is 91%, and the prevalence (P(U)) of drug use is 7%. Bayes' theorem states:
P(U|T+) = [P(T+|U) * P(U)] / [P(T+|U) * P(U) + P(T+|U') * P(U')]
First, we find the probability of a non-user testing positive, P(T+|U'), which is the complement of the specificity, 100% - 91% = 9%. The probability that an employee is a non-user is the complement of the drug use prevalence, P(U') = 100% - 7% = 93%.
Now we can calculate:
P(U|T+) = (0.98 * 0.07) / (0.98 * 0.07 + 0.09 * 0.93) = 0.0686 / (0.0686 + 0.0867) ≈ 0.0686 / 0.1553 ≈ 0.4419 or 44.19%
Even with a test that appears accurate, the probability that an employee who tested positive is actually a drug user is roughly 44.19%.