Hypergeometric distribution

While the binomial models $n$ independent trials (draws with replacement), the hypergeometric models $n$ draws without replacement from a finite urn. The key difference: without replacement the probabilities are not independent — each draw changes the composition of the urn.

Property — $X\sim\text{Hyp}(N,K,n)$

An urn contains $N$ objects, of which $K$ are “successes” and $N-K$ are “failures”. You draw $n$ objects without replacement. If $X$ is the number of successes drawn, then $P(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}, \qquad \max(0,\,n-(N-K))\le k\le \min(n,K).$ Mean and variance: $\mu = E[X] = n\cdot\frac{K}{N}, \qquad \sigma^2 = n\cdot\frac{K}{N}\cdot\frac{N-K}{N}\cdot\frac{N-n}{N-1}.$

Observation — Reading the formula

• $\binom{K}{k}$: ways of choosing the $k$ successes among the $K$ available.
• $\binom{N-K}{n-k}$: ways of choosing the $n-k$ failures among the $N-K$ available.
• $\binom{N}{n}$: total ways of choosing $n$ objects among $N$ (denominator = possible cases).

It is a pure combinatorial calculation: the ratio between favourable cases and possible cases (classical probability).

Links

Topics: Probability distributions
Concepts: Hypergeometric distribution · Expected value · Variance
Methods: Hypergeometric distribution
Skills: Combinatorics · Probability calculation