Hypergeometric Distribution — Mathematics Knowledge System

Overview

The hypergeometric distribution models the number of successes in a fixed number of draws from a finite population WITHOUT replacement, where the population contains a known number of successes and failures. If a population of size N contains K successes, and n items are drawn without replacement, the number of successes X follows a hypergeometric distribution with PMF P(X=k) = C(K,k)C(N-K,n-k)/C(N,n). This contrasts with the binomial distribution, which assumes draws are independent (with replacement); the hypergeometric distribution accounts for the fact that each draw changes the composition of the remaining population.

Intuition

Imagine an urn with a mix of red and blue balls, and you draw a handful without putting any back. Every ball you remove changes the odds for the next draw — pull out a red ball and there are fewer red balls left for the following draws. This is fundamentally different from flipping a coin repeatedly (the binomial setting), where every trial faces exactly the same odds. The hypergeometric distribution is the correct tool whenever sampling is 'used up' as you go, like dealing cards from a deck or auditing a batch of products without putting inspected ones back.

Formal Definition

Definition

For a population of size N with K successes, drawing n items without replacement, let X be the number of successes drawn:

P(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}

PMF

E[X] = n\cdot\frac{K}{N}

Mean

\operatorname{Var}(X) = n\cdot\frac{K}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1}

Variance

Worked Examples

Identify N=10, K=4 (red), n=3 draws, k=2 successes.
$P(X=2) = \frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}}$
Compute each binomial coefficient.
$\binom{4}{2}=6, \quad \binom{6}{1}=6, \quad \binom{10}{3}=120$
Combine the values.
$P(X=2) = \frac{6\times6}{120} = \frac{36}{120} = 0.3$

Answer: P(X=2) = 0.3.

Practice Problems

Difficulty 5/10

A deck of 52 cards has 4 aces. If 5 cards are drawn without replacement, find the probability of getting exactly 1 ace. (Use C(4,1)=4, C(48,4)=194580, C(52,5)=2598960.)

Difficulty 4/10

A batch of 20 items has 5 defective. If 4 items are inspected without replacement, find E[X], the expected number of defective items found.

Difficulty 5/10

A committee of 3 is chosen from a group of 5 men and 5 women (10 total), without replacement. Find the probability the committee has exactly 2 women. (Use C(5,2)=10, C(5,1)=5, C(10,3)=120.)

Quiz

Summary

The hypergeometric distribution counts successes when sampling n items without replacement from a finite population of N with K successes.
PMF: P(X=k) = C(K,k)C(N-K,n-k)/C(N,n); mean n(K/N).
Unlike the binomial, each draw changes the odds for subsequent draws since the population shrinks.

References

WebsiteWikipedia — Hypergeometric distribution