Suppose a finite population has size \(N\), with \(M\) “success” items and \(N-M\) “failure” items. If we draw \(n\) items without replacement, and let \(X\) be the number of successes in the sample, then:
\[ X \sim \text{Hypergeom}(N,M,n) \]
with probability mass function
\[ \text{P}(X = x) = \frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}} \]
for
\[ \max(0, n-(N-M)) \le x \le \min(n,M). \]
Parameter summary:
| Symbol | Meaning |
|---|---|
| \(N\) | population size |
| \(M\) | number of successes in the population |
| \(n\) | sample size (draws without replacement) |
| \(X\) | successes observed in the sample |
\[ \text{E}(X) = n\frac{M}{N} \]
\[ \text{V}(X) = n\frac{M}{N}\left(1-\frac{M}{N}\right)\frac{N-n}{N-1} \]
The factor \(\frac{N-n}{N-1}\) is the finite-population correction (FPC). It is strictly less than 1 when \(n>1\), which reduces variance relative to a binomial model with the same nominal success probability \(M/N\). Intuitively, sampling without replacement induces negative dependence among draws.
At least one mode is
\[ \text{Mo}(X)=\left\lfloor \frac{(n+1)(M+1)}{N+2}\right\rfloor. \]
If \(\frac{(n+1)(M+1)}{N+2}\) is an integer, two adjacent values can be modes.
No simple closed-form expression is available in general; the median is typically obtained numerically from the CDF.
Using the finite support of \(X\), the MGF can be written as
\[ M_X(t)=\sum_{x=\max(0,n-(N-M))}^{\min(n,M)} e^{tx}\,\text{P}(X=x). \]
\[ g_1=\frac{(N-2M)(N-2n)\sqrt{N-1}}{(N-2)\sqrt{nM(N-M)(N-n)}}. \]
\[ g_2 = 3 + \frac{(N-1)N^2\left[N(N+1)-6M(N-M)-6n(N-n)\right] + 6nM(N-M)(N-n)(5N-6)}{nM(N-M)(N-n)(N-2)(N-3)}. \]
The corresponding excess kurtosis is
\[ g_2 - 3 = \frac{(N-1)N^2\left[N(N+1)-6M(N-M)-6n(N-n)\right] + 6nM(N-M)(N-n)(5N-6)}{nM(N-M)(N-n)(N-2)(N-3)}. \]
The hypergeometric distribution is the default model for sampling without replacement:
The Hypergeometric Probabilities app is available in the handbook menu:
Distributions / Hypergeometric ProbabilitiesIt is also accessible directly at:
An organization has \(N = 1200\) procurement records for a quarter. Based on risk profiling, \(M = 90\) are flagged as high-risk. An internal audit samples \(n = 80\) records without replacement.
Let \(X\) be the number of high-risk records in the sample. A key escalation metric is:
\[ \text{P}(X \ge 10) \]
P(X >= 10) = 0.06893187
P(5 <= X <= 12) = 0.7297473 You can reproduce this setup with the preconfigured app below:
In a conservation study, a habitat has \(N=500\) tagged plants, of which \(M=80\) belong to a rare species.
A team samples \(n=40\) plants without replacement and records the number \(X\) of rare plants.
A monitoring question is:
\[ \text{P}(X \ge 10). \]