Hypergeometric distribution

Definition taught in Random variables and distributions, part of Probability

Hypergeometric distribution

Type Definition
Taught in Random variables and distributions
Part Probability
Worked examples 2

A hypergeometric count draws n items without replacement from a population of size N that contains r marked items. If X is the number of marked items in the draw, then

\mathbb{P}(X=x)=\frac{\binom{r}{x}\binom{N-r}{n-x}}{\binom{N}{n}},

for every integer x with \max(0,n-(N-r))\le x\le\min(n,r). Write p=r/N. Then \mathbb{E} X=np and

\operatorname{Var}(X)=np(1-p)\frac{N-n}{N-1}.

Examples

Aces in a hand

Worked example from Random variables and distributions

Five cards are dealt from a well-shuffled deck of 52.

Question. Find the distribution of the number of aces, and compare it with the binomial count for five draws with replacement.

Solution. The number of aces is hypergeometric with N=52, r=4 and n=5. For example,

\mathbb{P}(X=1)=\frac{\binom41\binom{48}4}{\binom{52}5}=\frac{4\times194580}{2598960}=0.2995.

The full distribution is 0.6588, 0.2995, 0.0399, 0.0017 and 0.00002 for 0 to 4 aces. The binomial count \operatorname{Bin}(5,1/13) gives 0.6702, 0.2792 and 0.0465 for 0 to 2. Both have mean 5/13=0.385, but without replacement each ace drawn makes the next less likely, so the hypergeometric count is less spread out.

Five draws without replacement

Question page: Five draws without replacement

Five numbers are drawn without replacement from \{1,\ldots,100\}. Find the probability that exactly three of them are even.

The draw is hypergeometric: 5 items from 100, of which 50 are even.

\mathbb{P}(X=3)=\frac{\binom{50}{3}\binom{50}{2}}{\binom{100}{5}}\approx 0.319.

Used in

See also

Random variable · Cumulative distribution function · Discrete distribution · Continuous distribution · Expectation · Expectation of a function · Variance · Probability generating function

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