Hypergeometric distribution

The hypergeometric distribution counts successes in a sample drawn without replacement from a finite population with a known number of successes.

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

pmf, for max(0, n − N + K) ≤ k ≤ min(n, K)

$N$: population size
$K$: number of successes in the population
$n$: sample size (drawn without replacement)
$k$: number of successes in the sample

\E(X) = n\frac{K}{N}, \qquad \Var(X) = n\frac{K}{N}\Big(1 - \frac{K}{N}\Big)\frac{N-n}{N-1}

Mean and variance

$\frac{N-n}{N-1}$: the finite population correction factor (squared FPC): sampling without replacement reduces the variance

When to use

Drawing without replacement from a small population: cards from a deck, a batch inspected for defectives, people selected for a committee. It is also the null distribution behind Fisher's exact test.

Assumptions

Every subset of size $n$ is equally likely (a simple random sample) and the population counts $N$ and $K$ are known.

How to interpret

Same mean as the binomial with $p = K/N$ , but a smaller variance, by the factor $(N-n)/(N-1)$ .
When $n$ is under about 10% of $N$ , the binomial gives nearly the same answers.

Common mistakes

Using the binomial when a large fraction of the population is sampled, which overstates the variability.
Mixing up $K$ (successes in the population) and $k$ (successes in the sample).

Example

A batch of $N = 20$ parts contains $K = 5$ faulty ones; $n = 4$ are inspected. $P(X = 1) = 0.4696$ ; $P(\text{at least one faulty}) = \text{sf}(0) = 0.7183$ . The binomial approximation ( $n = 4$ , $p = 0.25$ ) would give $0.6836$ , noticeably off because 20% of the batch is sampled.