Consider a company which has decided to implement a random drug test policy for

Question

Consider a company which has decided to implement a random drug test policy for its employees. In this case it is in the best interests of both the company and their employees that any such test is reliable. Suppose that 1% of all employees at the company are using illegal drugs and a further 0.05% use prescription drugs. Further suppose that the drug test put in place is 98% accurate in the sense that is indicates a positive test result for drugs 98% of the time when a person actually has illegal drugs in their system. However it also indicates a false positive 10% of the time when a person has prescription drugs in their system and 1.5% of the time if someone is not using any drugs. a) What is the probability that a randomly chosen employee tests positive for illegal drugs? (b) If a randomly chosen employee tests positive for illegal drugs, what is the probability that they are actually using illegal drugs? (c) What is the probability that a randomly chosen employee who tested negative for drugs is actually using illegal drugs? (d) In the light of all the above information, do you consider this to be a reliable test? Why or why not?

Explanation / Answer

Ans:

P(illegal drugs)=0.01

P(perscription drugs)=0.0005

P(not any drug)=1-0.01-0.0005=0.9895

P(positive/illegal drug)=0.98

P(negative/illegal drug)=1-0.98=0.02

P(positive/perscription drug)=0.1

P(positive/not any drug)=0.015

a)P(postive and illegal drugs)=P(positive/illegal drugs)*P(illegal drugs)

=0.98*0.01

=0.0098

b)P(illegal drugs/positive)=P(positive/illegal drugs)*P(illegal drugs)/P(positive)

P(positive)=P(positive/illlegal drugs)*P(illegal drugs)+P(postive/perscription drugs)*P(perscription drugs)+P(positive/not any drug)*P(not any drug)

P(positive)=0.98*0.01+0.1*0.0005+0.015*0.9895=0.025

P(illegal drugs/positive)=0.0098/0.025=0.392

c)P(illegal drugs/negative)=P(negative/illegal drugs)*P(illegal drugs)/P(negative)

P(negative)=1-0.025=0.975

P(illegal drugs/negative)=0.02*0.01/0.975=0.000205

d)As the probability of illegal drug given that test was negaive is very low,so it is reliable.

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