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Math Help - Probability Questions

  1. #1
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    Post Probability Questions

    I'm lost on how to answer these questions. I just don't know where to start. Please help!

    The probability of a french hen truly having citizenship is .81, find the probability that exactly 2 hens out of three have citizenship.

    and

    If there is an infinite number of calling birds, and the probability of a bird actually calling is .63, find the probability of finding the first calling bird on the third attempt.

    Please solve and then explain. Thanks!
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  2. #2
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    Quote Originally Posted by suchgreatheights View Post

    The probability of a french hen truly having citizenship is .81, find the probability that exactly 2 hens out of three have citizenship.
    Use the binomial distribution

    \displaystyle P(2) = ^3C_2(0.81)^2\times (1-0.81)^1=\dots
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by suchgreatheights View Post
    I'm lost on how to answer these questions. I just don't know where to start. Please help!

    The probability of a french hen truly having citizenship is .81, find the probability that exactly 2 hens out of three have citizenship.
    Let X be the random variable that represents the number of french hens truly having citizenship. Then X has a binomial distribution with n=3 and p=.81. Now what is P(X=2)?

    If there is an infinite number of calling birds, and the probability of a bird actually calling is .63, find the probability of finding the first calling bird on the third attempt.

    Please solve and then explain. Thanks!
    This one is a little more interesting because you're looking for the first success after so many trials. Hence, this one is modeled by a geometric distribution where p=.63. Now what is P(X=3)?

    I leave it to you to look up their corresponding pdfs and finish working out these problems. Can you proceed?
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  4. #4
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    Well, what would be setup of the geometric distribution for the second one?
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  5. #5
    MHF Contributor matheagle's Avatar
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    P(failure on first)P(failure on second)P(success on third attempt)

    =(.37)(.37)(.63)

    assuming indep between these selections
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