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Math Help - Probability of A given B is true

  1. #1
    Senior Member chella182's Avatar
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    Probability of A given B is true

    I'm pretty sure that's what kind of question this is anyway.

    A drugs test for athletes is 95% effective in detecting a steroid wen it is actually present. However, it also yields a false positive result for 1% of "clean" athletes tested. If 0.5% of athletes use the steroid, what is the probability that an athlete who tests positive did actually use the drug?
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  2. #2
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    Quote Originally Posted by chella182 View Post
    I'm pretty sure that's what kind of question this is anyway.

    A drugs test for athletes is 95% effective in detecting a steroid wen it is actually present. However, it also yields a false positive result for 1% of "clean" athletes tested. If 0.5% of athletes use the steroid, what is the probability that an athlete who tests positive did actually use the drug?
    Imagine that 100000 atheletes tested so that 0.005(100000)= 500 use the steroid and 99500 don't. Of the 500 who use the steroid 0.95(500)= 465 test positive. Of the 99500 who don't use the steroid 0.01(99500)= 995 also test positive. That gives a total of 995+ 465= 1465 positive tests of whom 465 actually use steroids. The probability that a positive test means the athelete did use the steroid is 465/1465.
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  3. #3
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    Hello, chella182!

    Yes, this is a Conditional Probability problem.


    A drugs test for athletes is 95% effective in detecting a steroid when it is actually present.
    However, it also yields a false positive result for 1% of "clean" athletes tested.
    If 0.5% of athletes use the steroid, what is the probability that an athlete
    who tests positive did actually use the drug?
    "0.5% of the atheletes use the drug": . P(\text{use}) \:=\:0.005,\;P(\text{clean}) \:=\:0.995


    "The test is 95% effective in detecting a steroid when it is actually present."
    If the athlete Uses the drug, the test will be Positive 95% of the time.
    . . That is: . P(\text{pos}\,|\,\text{use}) \:=\:0.95
    From Bayes' Theorem: . P(\text{pos}|\text{use}) \:=\:\frac{P(\text{pos}\wedge\text{use})}{P(\text{  use})} \:=\:0.95
    . . Then: . \frac{P(\text{pos}\wedge\text{use})}{0.005} \:=\:0.95 \quad\Rightarrow\quad P(\text{pos}\wedge\text{use}) \:=\:0.00475\;\;{\color{blue}[1]}



    "It yields a false positive result for 1% of 'clean' athletes tested."
    If an athlete is clean, the test will be positive 1% of the time.
    . . That is: . P(\text{pos}|\text{clean}) \:=\:0.01
    From Bayes' Theorem: . P(\text{pos}|\text{clean}) \:=\:\frac{P(\text{pos}\wedge\text{clean})}{P(\tex  t{clean})} \:=\:0.01

    . . Then: . \frac{P(\text{pos}\wedge\text{clean})}{0.995} \:=\:0.01 \quad\Rightarrow\quad P(\text{pos}\wedge\text{clean}) \:=\:0.00995 \;\;{\color{blue}[2]}


    From [1] and [2]: . P(\text{pos}) \:=\:0.00475 + 0.0095 \:=\:0.0147\;\;{\color{blue}[3]}



    "What is the probability that an athlete who tests positive did use the drug?"

    . . We want: . P(\text{use}|\text{pos}) \:=\:\frac{P(\text{use}\wedge\text{pos})}{P(\text{  pos})} <br />

    From [1] and [3]: . P(\text{use}|\text{pos}) \:=\:\frac{0.00475}{0.0147} \;\approx\;0.323 \;=\;32.3\%

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  4. #4
    Senior Member chella182's Avatar
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    Whether that's right or not, that deserves a thanks

    Although only 32% of people who test positive actually use?? That seems a little unrealisic to me...
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  5. #5
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    Quote Originally Posted by chella182 View Post
    Whether that's right or not, that deserves a thanks

    Although only 32% of people who test positive actually use?? That seems a little unrealisic to me...
    The answer is correct.

    The question has previously been asked and answered in this thread: http://www.mathhelpforum.com/math-he...obability.html

    (see post #9 -->)
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  6. #6
    Senior Member chella182's Avatar
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    Ahh Matthew posted it too haha! Sorry, he didn't tell me.
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