5.11: Hypergeometric Distribution

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:

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

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