Homework Answers
Consider: A1 = Person is a drug user
A2 = Person is not a drug user
B = Event = Positive drug test result
Required output is: P(A1|B)
Prior probabilities P(A1) = 0.023 and P(A2) = 1-0.023 = 0.977
P(B|A1) = 0.91
P(B|A2) = 0.09
Posterior probability:
P(A1|B) = [P(B|A1)*P(A1) ] / [P(B|A1)*P(A1) + P(B|A2)*P(A2) ]
= (0.91*0.023)/(0.91*0.023 + 0.09*0.977)
=0.192265295
=19.23%
Thus, the correct answer is 19.23%
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