Bayes Theorem probability of steroids

[SammC]

Homework Statement

When the test for steroids is given to soccer players, 98% of players taking steroids test positive and 0.5% of the players not taking steroids test positive. Suppose that 5% of soccer players take steroids, what is the probability that a player who tests positive takes steroids.


2. Homework Equations and attempt at solution
p: tests positive
s: uses steroids

pr(s) = .05
pr(P|S) = .98

inclusion exclusion pr(p) = Pr(P|S)*Pr(S) + Pr(P|not S) * Pr(not S) = .98*.5 + .005 *.95 = .05375

Bayes Theorem: [Pr(S | P) = Pr(P | S) * Pr(S)]/Pr(p)

plugging in...
Pr(S | P) = .91

My question: Is 91% a plausible answer to this problem? Its the first one I've ever done using Bayes rule.

 

[HallsofIvy]

Imagine that there are 100000 people tested. 5% of them, 5000, use steroids, the other 100000- 5000= 95000 do not. Of those 5000 who use steroids, 98%= 4900 test positive. Of the 95000 who do not use steriods, .5% of them= 475 also test positive.

So you have a pool of 4900+ 475= 5375 who test positive and 4900 of those take steroids: 4900/5375= 0.91. Yes, 91% is a plausible answer.

 

[SammC]

Thanks Much