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Bayes Theorem probability of steroids

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    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. Relevant 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.
     
  2. jcsd
  3. Mar 3, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  4. Mar 3, 2009 #3
    Thanks Much
     
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