Question

According to the guidelines of the World Anti-Doping Agency (WADA), athletes are subject to testing both ‘in-competition’ and ‘out-of-competition’ for prohibited substances. Suppose that the tests for prohibited substances can identify a substance abuse 96% of the time. However, 2% of the time the test indicates a positive result although the athlete is not a prohibited substance user. If 5% of all athletes are users of prohibited substances, what is the probability that an athlete who has tested positive for a prohibited substance is not really a prohibited substance user? Round your answers to the nearest ten-thousandth (4 decimals).

Answer 1

We are given here that:
P( positive | substance ) = 0.96
P( negative | substance ) = 1 - 0.96 = 0.04

P( positive | no substance ) = 0.02,
P( negative | no substance ) =1 - 0.02 = 0.98

Also, we are given here that:
P( substance ) = 0.05, and therefore P( no substance ) = 0.95

Using law of total probability, we have here:
P( positive ) = P( positive | substance ) P( substance ) + P( positive | no substance ) P( no substance )

P( positive ) = 0.96*0.05 + 0.02*0.95 = 0.067

probability that an athlete who has tested positive for a prohibited substance is not really a prohibited substance user is computed using Bayes theorem here as:

P( no substance | positive ) = P( positive | no substance ) P( no substance ) / P( positive )

= 0.02*0.95 / 0.067

= 0.2836

Therefore 0.2836 is the required probability here.