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

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

So, how do you get from this to

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

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

Re: Hypergeometric probabilities sum to one?

Quote:

Originally Posted by

**Mathsdog** 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

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

So, how do you get from this to

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

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

Re: Hypergeometric probabilities sum to one?

That's last post by Monsieur Fantastique; about 3.5 years ago.

Who scared him away?

Looks like nobody cares anyway :)