A company's management is worried about employee drug use and is instituting a policy of mandatory drug tests. 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 a 95% chance that he or she will test negative. Individuals who test positive are fired. We do not 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 test positive, what fraction are in fact not drug users?

a) 0.0625 (6.25%)

b) 0.047 (4.7%)

Step-by-step explanation:

A) denoting the event T= the test is positive, then the probability is

P(T)= probability that the employee is drug user * probability that the test is positive if he is a drug user + probability that the employee is not a drug user * probability that the test is positive if he is not a drug user =

P(T)= p * 0.85 + (1-p)* (1-0.95) = 0.10

0.85*p - 0.05*p = 0.10 - 0.05

p= 0.05/0.8 = 0.0625

then the proportion p of employees that are drug users is 0.0625 (6.25%)

B) the fraction that are not drug users but the test was positive is

(1-p)*(1-0.95)  = 0.9395*0.05 = 0.047 (4.7%)

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: