# 5.11: Hypergeometric Distribution

• • Contributed by David Lane
• Associate Professor (Psychology, Statistics, and Management) at Rice University

Learning Objectives

• To study the use of hypergeometric distribution

The hypergeometric distribution is used to calculate probabilities when sampling without replacement. For example, suppose you first randomly sample one card from a deck of $$52$$. Then, without putting the card back in the deck you sample a second and then (again without replacing cards) a third. Given this sampling procedure, what is the probability that exactly two of the sampled cards will be aces ($$4$$ of the $$52$$ cards in the deck are aces). You can calculate this probability using the following formula based on the hypergeometric distribution:

$p =\dfrac{ (_{k}C_{x})(_{(n-k)}C_{(n-x)}) }{ _{n}C_{n}}$

where

• $$k$$ is the number of "successes" in the population
• $$x$$ is the number of "successes" in the sample
• $$N$$ is the size of the population
• $$n$$ is the number sampled
• $$p$$ is the probability of obtaining exactly $$x$$ successes
• $$_kC_x$$ is the number of combinations of $$k$$ things taken $$x$$ at a time

In this example, $$k = 4$$ because there are four aces in the deck, $$x = 2$$ because the problem asks about the probability of getting two aces, $$N = 52$$ because there are $$52$$ cards in a deck, and $$n = 3$$ because $$3$$ cards were sampled. Therefore,

\begin{align} p &=\dfrac{(_4C_2) (_{(52-4)}C_{(3-2})}{_{52}C_3} \\[5pt] &= \dfrac{\dfrac{4!}{2!2!}\dfrac{48!}{47!1!}}{\dfrac{52!}{49!3!}} = 0.013 \end{align}

The mean and standard deviation of the hypergeometric distribution are:

$mean = \dfrac{n\,k}{N}$

$\sigma_{hypergeometric} = \sqrt{\dfrac{n\,k(N-k)(N-m)}{N^2(N-1)}}$

## Contributor

• Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.