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Math Help - Urn/ball probability question

  1. #1
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    Urn/ball probability question

    suppose there is an urn containing c yellow balls and d green balls. we draw k balls without replacement, from the urn. find the expected number of yellow balls drawn.

    I wrote out my random variables: X_1 + X_2 + ... + X_c but I'm having trouble finding the probabilities
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  2. #2
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    P(X=n)=\displaystyle \frac{\binom{c}{n}\binom{d}{k-n}}{\binom{c+d}{k}},~0\le n\le k
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  3. #3
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    thanks!

    does this probability hold for all the random variables even though the balls are taken from the urn without replacement?
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  4. #4
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    Quote Originally Posted by morganfor View Post
    does this probability hold for all the random variables even though the balls are taken from the urn without replacement?
    There are some restrictions.
    0\le n \le c,~k\le c+d.

    So take care finding expection if c<k.
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  5. #5
    MHF Contributor matheagle's Avatar
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    The mean of a hypergeometric is the same as a binomial, the variances are different.

    Hypergeometric distribution - Wikipedia, the free encyclopedia

    The mean here should be 'np'=k{c\over c+d}={kc\over c+d}
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