Bayes Theorem Drug Testing > 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? Here is what I have so far: 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 I want to find P(Y|X) = $\frac{P(X|Y) P(Y)}{P(X)}$ however, I'm having trouble finding P(Y). What I know I should be using is P(Y) = P(Y|X) P(X) + P(Y|notX) P(notX) but I get results over 1 when I plug this formula :( What am I doing wrong? Thanks in advance for your help!

Hint: 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 (as you have done $$P(Y)=P(Y|X)P(X)+P(Y|X^c)P(X^c)=0.98\cdot0.02+0.03\cdot0.98=0.049$$