A large company gives a new employee a drug test. The False-Positive rate is 3% and the False-Negative rate is 2%. In addition, 2% of the population use the drug. The employee tests positive for the drug. What is the probability the employee uses the drug?

Yazmin Sims 2022-10-27 Answered
A large company gives a new employee a drug test. The False-Positive rate is 3% and the False-Negative rate is 2%. In addition, 2% of the population use the drug. The employee tests positive for the drug. What is the probability the employee uses the drug?
What I try:
X={employee uses drugs}
Y={employee tests positive in the drug test}
P(X|Y) = 0.97
P(notX|Y) = 0.03
P(X|notY) = 0.02
P(notX|notY) = 0.98
P(X) = 0.02
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Answers (1)

Adenomacu
Answered 2022-10-28 Author has 13 answers
You have confused throughout your solution (which however is methodologically correct) the roles of 𝑋 and 𝑌. You know that
1. P ( Y | X ) = 0.98 ( = 1 False Negative) and
2. P ( Y | X c ) = 0.03
Accordingly you have confused P ( Y | X ) with the actual required probability which is P ( X | Y ). For the calculation of P ( Y ) which will come in the denominator of the Bayes fromula you should use the Law of total probability
P ( Y ) = P ( Y | X ) P ( X ) + P ( Y | X c ) P ( X c ) = 0.98 0.02 + 0.03 0.98 = 0.049
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