Question

A company screens job applicants for illegal drug use at a certain stage in their hiring process. The specific test they use has a false positive rate of 2% and false negative rate of 1%. Suppose that 5% of all their applicants are actually using illegal drugs. One employee is randomly selected and he was tested positive.

a) What are the prior probabilties that the job applicant use or does not use illegal drugs?

b) What are the revised probabilities that the job applicant use or does not use illegal drugs?

c) Based on your analysis, what is the overall percentage of job applicants that test positive?

Answer #1

a) Prior probability of a job applicant using drugs = 0.05

Prior Probability of a job applicant not using drugs = 0.95

b) Revised Probability of a job applicant using drugs given that he is tested positive = P(positive and Uses drugs)/ P(Positive)

= (0.05*0.99) / (0.05*0.99 + 0.95*0.02)

= **0.7226**

Revised Probability of a job applicant not using drugs given that he is tested positive = P(positive and not uses drugs)/ P(Positive)

= (0.95*0.02) / (0.05*0.99 + 0.95*0.02)

= **0.2774**

c) Percentage of job applicant who are tested positive = P(use drugs and +) + P(Not use drugs and +)

= 0.05*0.99 + 0.95*0.02 = **0.0685**

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