hypergeometric distribution

hypergeometric distribution has the following properties:
 The mean of the distribution is equal to n * k / N .
 The variance is n * k * ( N - k ) * ( N - n ) / [ N2 * ( N - 1 ) ] .

Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)?
Solution: This is a hypergeometric experiment in which we know the following:
 N = 52; since there are 52 cards in a deck.
 k = 26; since there are 26 red cards in a deck.
 n = 5; since we randomly select 5 cards from the deck.
 x = 2; since 2 of the cards we select are red.
We plug these values into the hypergeometric formula as follows:
h(x; N, n, k) = [ kCx ] [ N-kCn-x ] / [ NCn ]
h(2; 52, 5, 26) = [ 26C2 ] [ 26C3 ] / [ 52C5 ]
h(2; 52, 5, 26) = [ 325 ] [ 2600 ] / [ 2,598,960 ] = 0.32513
Thus, the probability of randomly selecting 2 red cards is 0.32513.
Cumulative Hypergeometric Probability
A cumulative hypergeometric probability refers to the probability that the hypergeometric random variable is greater than or equal to some specified lower limit and less than or equal to some specified upper limit.
For example, suppose we randomly select five cards from an ordinary deck of playing cards. We might be interested in the cumulative hypergeometric probability of obtaining 2 or fewer hearts. This would be the probability of obtaining 0 hearts plus the probability of obtaining 1 heart plus the probability of obtaining 2 hearts, as shown in the example below.
Example 1

Suppose we select 5 cards from an ordinary deck of playing cards. What is the probability of obtaining 2 or fewer hearts?
Solution: This is a hypergeometric experiment in which we know the following:
 N = 52; since there are 52 cards in a deck.
 k = 13; since there are 13 hearts in a deck.
 n = 5; since we randomly select 5 cards from the deck.
 x = 0 to 2; since our selection includes 0, 1, or 2 hearts.
We plug these values into the hypergeometric formula as follows:
h(x < x; N, n, k) = h(x < 2; 52, 5, 13)
h(x < 2; 52, 5, 13) = h(x = 0; 52, 5, 13) + h(x = 1; 52, 5, 13) + h(x = 2; 52, 5, 13)
h(x < 2; 52, 5, 13) = [ (13C0) (39C5) / (52C5) ] + [ (13C1) (39C4) / (52C5) ] + [ (13C2) (39C3) / (52C5) ]
h(x < 2; 52, 5, 13) = [ (1)(575,757)/(2,598,960) ] + [ (13)(82,251)/(270,725) ] + [ (78)(9139)/(22,100) ]
h(x < 2; 52, 5, 13) = [ 0.2215 ] + [ 0.4114 ] + [ 0.2743 ]
h(x < 2; 52, 5, 13) = 0.9072
Thus, the probability of randomly selecting at most 2 hearts is 0.9072.