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Hypergeometric Distribution

Definition

Let a population contain NN objects, of which KK are successes (0≤K≤N0\le K\le N), and suppose nn objects (0≤n≤N0\le n\le N) are drawn without replacement. A random variable XX has a hypergeometric distribution if XX counts the number of successes in the draw, with

P(X=k)=(Kk)(N−Kn−k)(Nn),max⁡(0, n−N+K)≤k≤min⁡(n,K),\begin{gathered} \mathbb{P}(X=k)=\frac{\binom Kk\binom{N-K}{n-k}}{\binom Nn},\\ \max(0,\,n-N+K)\le k\le\min(n,K), \end{gathered}

and P(X=k)=0\mathbb{P}(X=k)=0 for all other kk.


Interpretation

A hypergeometric distribution counts successes when sampling without replacement. The trials are dependent because each draw changes what remains.

Probability bars for the number of successes among 6 objects drawn without replacement from 20 objects of which 8 are successes, peaking at 2 with mean 2.4.

The diagram counts successes in a sample of n=6n=6 objects drawn without replacement from N=20N=20 objects, of which K=8K=8 are successes. The PMF is P(X=k)=(8k)(126−k)/(206)\mathbb{P}(X=k)=\binom8k\binom{12}{6-k}/\binom{20}6 for 0≤k≤60\le k\le6. Its mean is 6⋅820=2.46\cdot\tfrac{8}{20}=2.4, and its variance 1.44⋅1419≈1.0611.44\cdot\tfrac{14}{19}\approx1.061 is smaller than the variance 6(0.4)(0.6)=1.446(0.4)(0.6)=1.44 of Bin⁡(6,0.4)\operatorname{Bin}(6,0.4), the corresponding model with replacement.

Change NN, KK and nn: drawing without replacement gives mean nK/NnK/N, like the binomial, but a smaller variance because each draw changes what remains.

A combination counts an unordered selection without replacement. Counting the chosen successes and failures separately and dividing by all samples gives the hypergeometric probabilities.


Key Results

  • The hypergeometric distribution models sampling without replacement.
  • Unlike the binomial distribution, the trials are dependent.
  • Its mean is nK/NnK/N when KK successes are in a population of size NN and nn items are drawn.
  • Its variance is

    Var⁡(X)=nKN(1−KN)N−nN−1.\operatorname{Var}(X)=n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}.

    This is the binomial variance with p=K/Np=K/N times the finite population correction N−nN−1\frac{N-n}{N-1}, which is below 11 when n≥2n\ge2 and close to 11 when the population is much larger than the sample.


Examples

  • The number of aces in a five-card hand has a hypergeometric distribution.
  • The number of defective items in a sample drawn without replacement from a shipment is hypergeometric.
  • The number of red balls drawn from an urn without replacement is hypergeometric.

Prerequisites

Sources

Keywords: hypergeometric random variable, sampling without replacement

Last updated: 20 October 2026 03:02