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Hypergeometric distribution

The number of successes in a sample drawn WITHOUT replacement from a finite population.

Parameters and support

$N$: population size — $N \in {1, 2, \dots}$
$K$: number of successes in the population — $0 \leq K \leq N$
$n$: sample size drawn without replacement — $0 \leq n \leq N$
Support: $max (0, n - (N - K)) \leq x \leq min (n, K)$

The formulas

p(x)

p (x) = \frac{( x K ) ( n - x N - K )}{( n N )}

F(x)

No elementary closed form.

Summed term by term; there is no closed form and none is expected on the exam.

Mean

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

Variance

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

MGF

Not examinable for this distribution.

The MGF is a hypergeometric series with no useful closed form. It is not examinable.

Memory hook. Same mean as a binomial with p = K/N. The variance is the binomial variance times the finite-population correction (N − n)/(N − 1), which is always ≤ 1 — sampling without replacement is LESS variable.

Where the moments come from

Choose $x$ of the $K$ successes and $n - x$ of the $N - K$ failures, out of all $(n N)$ equally likely samples.
Write $X = \sum_{k = 1}^{N} I_{k}$ where $I_{k}$ indicates that population member $k$ is both a success and selected.
Each indicator has $E [I_{k}] = \frac{K}{N} \cdot \frac{n}{N}$ -style symmetry giving $E [X] = n \frac{K}{N}$ — identical to the binomial.
The indicators are NEGATIVELY correlated (drawing one success leaves fewer), and that covariance produces the factor $\frac{N - n}{N - 1}$ .
As $N \to \infty$ with $K / N \to p$ the correction tends to 1 and the hypergeometric becomes binomial.

Worked example

A file drawer holds 20 claims, 6 of which are fraudulent. An auditor samples 5 claims without replacement. Find the probability that exactly 2 sampled claims are fraudulent.

Sampling is without replacement from a finite population, so this is hypergeometric, not binomial.
$P (X = 2) = \frac{( 2 6 ) ( 3 14 )}{( 5 20 )}$ .
$(2 6) = 15$ , $(3 14) = 364$ , $(5 20) = 15504$ .
$P (X = 2) = \frac{15 \times 364}{15504} = \frac{5460}{15504} = 0.3522$ .

Answer: 0.3522

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run — summed over the support for a discrete distribution, integrated by quadrature for a continuous one — and compared with the closed forms printed here. A typo on this page fails the build.

Traps

Treating the draws as independent and using the binomial — the giveaway phrase is 'without replacement'.
Dropping the (N − n)/(N − 1) correction from the variance.
Forgetting that the support is truncated when n > N − K.

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.

Hypergeometric distribution

Parameters and support

The formulas

Where the moments come from

Worked example

Traps

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