suppose that 1% of the employees of a certain company use illegal drugs. This co

Question

suppose that 1% of the employees of a certain company use illegal drugs. This company performs random drug tests that return positive results 99% of the time if the person is a drug user. However, it also has a 2% false positve rate. The results of the drug test are known to be independent from test to test for a given person.

(a) Steve, an employee at the company, has a positve test. What is the probability he is a drug user?

(b) Knowing he failed his first test, what is the probability that Steve will fail his next drug test?

(c) Steve just failed his second drug test. Now what is the probability he is a drug user?

Explanation / Answer

Ans:

Given that

P(drug)=0.01

P(not drug)=1-0.01=0.99

P(positive/drug)=0.99

P(negative/drug)=1-0.99=0.01

P(postive/not drug)=0.02

P(negative/not drug)=1-0.02=0.98

P(positive)=P(positive/drug)*P(drug)+P(positive/not drug)*P(not drug)

=0.99*0.01+0.02*0.99=0.0297

a)P(drug/positive)=P(positive/drug)*P(drug)/P(positive)

=0.99*0.01/0.0297=0.333

b)As,tests are independent,so

P(failed second test)=P(negative)=1-0.0297=0.9703

c)P(drug/negative)=P(negative/drug)*P(drug)/P(negative)=0.01*0.01/(1-0.0297)=0.0001

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