Question

A drug test for athletes has a 5% false positive rate and a 10% false negative rate. Of the athletes tested, 4% have actually been using the prohibited drug. If a randomly chosen athlete tests positive, what is the probability that the prohibited drug has been used

Answer 1

Answer:

0.4286 or 42.86%

Step-by-step explanation:

The probability that an athlete has used drugs and tests positive is given by the probability that he had been using drugs (4%) multiplied by the probability of a true positive (1 - false negative):

$P(D\ and\ +) = 0.04*(1-0.10)=0.036$

The probability that an athlete has NOT used drugs and tests positive is given by the probability that he had not been using drugs (100% - 4%) multiplied by the probability of a false positive (5%):

$P(N\ and\ +) = (1-0.04)*0.05=0.048$

Finally,  the probability that the prohibited drug has been used given that an athlete tested positive is:

$P(D|+)=\frac{P(D\ and\ +)}{P(D\ and\ +) \ +\ P(N\ and\ +)} \\P(D|+)=\frac{0.036}{0.036+0.048}=0.4286$

The probability is 0.4286 or 42.86%.