Editing Hypergeometric distribution (section)

===Combinatorial identities===

As required, we have

:<math display="block"> \sum_{0\leq k\leq \textrm{min}(n,K)} { {K \choose k} { N-K \choose n-k} \over {N \choose n} } 
= 1,</math>

which essentially follows from [[Vandermonde's identity]] from [[combinatorics]].

Also note that

:<math> {{K \choose k} {N-K \choose n-k}\over {N \choose n}} = 
 {{{n \choose k} {{N-n} \choose {K-k}}} \over {N \choose K}};</math>

This identity can be shown by expressing the binomial coefficients in terms of factorials and rearranging the latter. Additionally, it
follows from the symmetry of the problem, described in two different but interchangeable ways. 

For example, consider two rounds of drawing without replacement. In the first round, <math>K</math> out of <math>N</math> neutral marbles are drawn from an urn without replacement and coloured green. Then the colored marbles are put back. In the second round, <math>n</math> marbles are drawn without replacement and colored red. Then, the number of marbles with both colors on them (that is, the number of marbles that have been drawn twice) has the hypergeometric distribution. The symmetry in <math>K</math> and <math>n</math> stems from the fact that the two rounds are independent, and one could have started by drawing <math>n</math> balls and colouring them red first.

''Note that we are interested in the probability of <math>k</math> successes in <math>n</math> draws '''without replacement''', since the probability of success on each trial is not the same, as the size of the remaining population changes as we remove each marble. Keep in mind not to confuse with the [[binomial distribution]], which describes the probability of <math>k</math> successes in <math>n</math> draws '''with replacement.'''''