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Thread: Probability with sampling question

  1. #16
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    Re: Probability with sampling question




    I got this:
    A: 1st person in sample gets drug A
    B: 2nd person in sample gets drug A
    C: 3rd person in sample gets drug A
    D: 4th person in sample gets drug A

    P(A)*P(B|A)*P(C|A and B)*P(D | A and B and C)

    The problem is that I don`t have the joint probabilities A and B, A and B and C.

    I am assuming sampling without replacement, so there is dependence.

  2. #17
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    Re: Probability with sampling question

    Can you figure out P(A)?

    What do you think P(B|A) is? Just think about what it means. It means that you already picked somebody that had drug A (so there are only 129 people you can choose from now and out of the original 25 patients that had drug A there are now only 24 of them that you could choose).
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  3. #18
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    Re: Probability with sampling question


    P(A)*P(B|A)*P(C|A and B)*P(D | A and B and C)
    =(25/130)*(24/129)*(23/128)*(22/127)

    ???

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