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Hypergeometric distribution

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

Parameters and support

NN
population size - N∈{1,2,… }N \in \{1,2,\dots\}
KK
number of successes in the population - 0≤K≤N0 \le K \le N
nn
sample size drawn without replacement - 0≤n≤N0 \le n \le N
Support
max⁡(0, n−(N−K))≤x≤min⁡(n, K)\max(0,\, n-(N-K)) \le x \le \min(n,\, K)

The formulas

p(x)
p(x)=(Kx)(N−Kn−x)(Nn)p(x) = \dfrac{\dbinom{K}{x}\dbinom{N-K}{n-x}}{\dbinom{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]=nKNE[X] = n\dfrac{K}{N}
Variance
Var⁡(X)=nKN(1−KN)N−nN−1\operatorname{Var}(X) = n\dfrac{K}{N}\left(1-\dfrac{K}{N}\right)\dfrac{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

  1. Choose xx of the KK successes and n−xn-x of the N−KN-K failures, out of all (Nn)\binom{N}{n} equally likely samples.
  2. Write X=∑k=1NIkX = \sum_{k=1}^{N} I_k where IkI_k indicates that population member kk is both a success and selected.
  3. Each indicator has E[Ik]=KN⋅nNE[I_k] = \frac{K}{N}\cdot\frac{n}{N}-style symmetry giving E[X]=nKNE[X] = n\frac{K}{N} - identical to the binomial.
  4. The indicators are NEGATIVELY correlated (drawing one success leaves fewer), and that covariance produces the factor N−nN−1\frac{N-n}{N-1}.
  5. As N→∞N \to \infty with K/N→pK/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.

  1. Sampling is without replacement from a finite population, so this is hypergeometric, not binomial.
  2. P(X=2)=(62)(143)(205)P(X = 2) = \dfrac{\binom{6}{2}\binom{14}{3}}{\binom{20}{5}}.
  3. (62)=15\binom{6}{2} = 15, (143)=364\binom{14}{3} = 364, (205)=15504\binom{20}{5} = 15504.
  4. P(X=2)=15×36415504=546015504=0.3522P(X = 2) = \dfrac{15 \times 364}{15504} = \dfrac{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 that test run.

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.

Related

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.