The Hypergeometric Distribution
Suppose you have a full deck of 52 cards and you randomly draw 6 cards from the deck without replacing any of the cards. What is the probability that 2 of the cards are kings?
The hypergeometric distribution is a discrete probability distribution used to express probabilities when sampling without replacement. More specifically, a hypergeometric distribution describes the probabilities of k successes in n draws without replacement from a finite population of size N which contains exactly K successful states.
A hypergeometric random variable X has the following probability distribution:
PX=k = KkN−Kn−kNn,
where N is the population size, K is the number of successful states in the population, n is the number of draws or events, and k is the number of successes.
If N is the population size, K is the number of successful states in the population, n is the number of draws, and k is the number of success then:
The probability mass function.
The expected value of a random variable.
n KN N−KN N−nN−1
Represented by the symbol σ2, representing how much variation or spread exists from the mean value.
Suppose you have a full deck of cards and you randomly choose 6 cards without replacement. What is the probability that you will choose exactly 2 kings?
In this example, drawing a king from the deck of cards can be considered a success, meaning that since there are 4 kings in the deck of cards, there are 4 successful states.
Let population size, N, be 52, number of successful states (the number of kings in the deck of cards), K, be 4, number of draws, n, be 6, and number of successes, k, be 2.
fX =4 = PX=k = 4252−46−2526 = 0.057346.
The probability of randomly choosing 6 cards without replacement and obtaining exactly 2 kings is 0.057346 or 5.7%.
Change the population size N, the number of successful states in the population K, the number of draws n, and the number of successes k to observe the change in the distribution:
size of the population (N) =
# of successes in population (K) =
# of draws (n) =
# of successes (k) =
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