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) = 0.97P(notX|Y) = 0.03P(X|notY) = 0.02P(notX|notY) = 0.98P(X) = 0.02

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Adenomacu · Accepted Answer

You have confused throughout your solution (which however is methodologically correct) the roles of 𝑋 and 𝑌. You know that1.   P  (  Y      |    X  )  =  0.98    (  =  1  − False Negative) and2.   P  (  Y      |        X    c    )  =  0.03Accordingly 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

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?

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