Suppose a certain drug test is 94% sensitive, that is, the test will correctly i

Question

Suppose a certain drug test is 94% sensitive, that is, the test will correctly identify a drug user as testing positive 94% of the time, and 94% specific, that is, the test will correctly identify a non-user as testing negative 94% of the time. Suppose a corporation decides to test its employees for drug use, and that only 0.6% of the employees actually use the drug. What is the probability that, given a positive drug test, an employee is actually a drug user? Let DD stand for being a drug user and NN indicate being a non-user. Let be the event of a positive drug test and be the event of a negative drug test. (b) P(N)=P(N)= Answer Incorrect (d) P(|N)=P(|N)= Answer Correct (e) P()=P()= Answer Incorrect (f) P(D|)

Explanation / Answer

Ans:

Given that

P(positive/use drug)=0.94

P(negative/use drug)=1-0.94=0.06

P(negative/not use drug)=0.94

P(positive/not use drug)=1-0.94=0.06

P(use drug)=0.006

P(not use drug)=1-0.006=0.994

We have to find

P(use drug/postive)=P(positive/use drug)*P(use drug)/[P(positive/use drug)*P(use drug)+P(positive/not use drug)*P(not use drug)]

=0.94*0.006/[0.94*0.006+0.06*0.994]

=0.00564/[0.00564+0.05964]

=0.00564/0.06528

=0.0864

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