Question

Problem 4.

(problem 5.23 page 74) A company‘s management is worried about employee drug

use and is instituting a policy of mandatory drug test. If an employee is a drug user, there is an 85%

chance that he or she will test positive. If the employee is not a drug user, there is 95% chance that

he or she will test negative. Individuals who test positive are fired. We don‘t know the fraction of

employees who are drug users. We do know, however that 10% of all employees tested positive.

a. What fraction of employees are drug users?

b. Of those who tested positive, what fraction are in fact not drug users?

Answer 1

P(tested positive | drug user) = 0.85

P(tested negative | not a drug user) = 0.95

P(tested positive | not a drug user) = 1 - P(tested negative | not a drug user) = 1 - 0.95 = 0.05

P(tested positive) = 0.1

a) P(tested positive) = P(tested positive | drug user) * P(drug user) + P(tested positive | not a drug user) * P(not a drug user)

or, 0.1 = 0.85 * P(drug user) + 0.05 * (1 - P(drug user))

or, 0.1 = 0.8 * P(drug user) + 0.05

or, P(drug user) = 0.0625

b) P(not drug user | tested positive) = P(tested positive | not a drug user) * P(not a drug user) / P(tested positive)

                                                         = 0.05 * (1 - 0.0625) / 0.1

                                                         = 0.46875