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Math Help - Hypergeometric probabilities sum to one?

  1. #1
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    Hypergeometric probabilities sum to one?

    Hi, Guess this should be easy, but could do with a pointer. Show directly that the set of probabilities associated with the hypergeometric distribution sum to 1. Hint: Expand the identity

    (1+ \mu)^N = (1+ \mu)^r (1+\mu)^{N-r}

    So, how do you get from this to

    \Sigma_{k=0}^r {{ \binom{r}{k} \cdot \binom{w}{n-k}} \over  \binom{N}{n} }} = 1

    which I take to be the formal statement of the verbal statement above, right? (i.e. the set of probabilities associated with the hypergeometric distribution sum to 1 (for a given r, w and N and where r + w = n)).

    As always, much appreciated and thanks in advance for any insights.
    Ta, MD
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  2. #2
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    Re: Hypergeometric probabilities sum to one?

    Quote Originally Posted by Mathsdog View Post
    Hi, Guess this should be easy, but could do with a pointer. Show directly that the set of probabilities associated with the hypergeometric distribution sum to 1. Hint: Expand the identity

    (1+ \mu)^N = (1+ \mu)^r (1+\mu)^{N-r}

    So, how do you get from this to

    \Sigma_{k=0}^r {{ \binom{r}{k} \cdot \binom{w}{n-k}} \over \binom{N}{n} }} = 1

    which I take to be the formal statement of the verbal statement above, right? (i.e. the set of probabilities associated with the hypergeometric distribution sum to 1 (for a given r, w and N and where r + w = n)).

    As always, much appreciated and thanks in advance for any insights.
    Ta, MD
    As is usually the case, all the necessary clues can be found using your search engine of choice:

    Hypergeometric distribution - Wikipedia, the free encyclopedia

    Vandermonde's identity - Wikipedia, the free encyclopedia
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