# 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?

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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You have confused throughout your solution (which however is methodologically correct) the roles of 𝑋 and 𝑌. You know that
1. $P\left(Y|X\right)=0.98\phantom{\rule{thinmathspace}{0ex}}\left(=1-$ False Negative) and
2. $P\left(Y|{X}^{c}\right)=0.03$
Accordingly you have confused $P\left(Y|X\right)$ with the actual required probability which is $P\left(X|Y\right)$. For the calculation of $P\left(Y\right)$ which will come in the denominator of the Bayes fromula you should use the Law of total probability
$P\left(Y\right)=P\left(Y|X\right)P\left(X\right)+P\left(Y|{X}^{c}\right)P\left({X}^{c}\right)=0.98\cdot 0.02+0.03\cdot 0.98=0.049$
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