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Thread: Conditional Probability

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    Conditional Probability




    http://i.imgur.com/9UELFY5.png
    Here is the problem in question. I have the answers for both of these already,but I am not sure I have 2b correct.

    http://i.imgur.com/9NmUgD8.png

    Here are the probability for 2a For 2b I just thought you had to take 95/(95+15) to get the probability, is that right or do I need to include the 13% somehow?

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    Re: Conditional Probability

    The 13% is relevant to the problem. If you don't use it then you aren't going about the problem correctly.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Conditional Probability

    Quote Originally Posted by Dason View Post
    The 13% is relevant to the problem. If you don't use it then you aren't going about the problem correctly.
    So, should I organize it something like this? P(drugs)=P(drugs|positive)P(positive)+P(drugs|negative)P(negative) and then rearrange to find P(drugs|positive) ?

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    Re: Conditional Probability

    That's not the conventional way we would do it but it should give a correct answer. Note that you'll need to figure out P(positive) and P(negative) to go this route.
    I don't have emotions and sometimes that makes me very sad.

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    Re: Conditional Probability

    Quote Originally Posted by Dason View Post
    That's not the conventional way we would do it but it should give a correct answer. Note that you'll need to figure out P(positive) and P(negative) to go this route.
    Oh there is another route? My lecture notes only have this route. Welp

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    Re: Conditional Probability


    After doing some searching:
    The probability that someone tests positive is (0.95)(0.13) + (0.15)(0.87).
    So I want want (0.95)(0.13)/((0.95)(0.13) + (0.15)(0.87)), which gives me 48.62% someone is doping?

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