Hypergeometric distribution ($K\sim \mathrm{Hypergeom}(n,c,r)$)

The hypergeometric distribution counts marked objects in a uniformly chosen sample of fixed size from a finite population, using sampling without replacement. Let $n\ge 1$ be the population size, let $0\le c\le n$ objects be marked, and sample $0\le r\le n$ distinct objects. If $K$ counts the marked objects selected, its probability mass function isThe bounds describe the support of a probability distribution and $k$ must be an integer; the mass is zero elsewhere. The denominator counts all size-$r$ subsets, while the numerator counts choices of $k$ marked and $r-k$ unmarked objects using binomial coefficients. This counting argument requires every size-$r$ sample to have the same probability.^[ref]

For a constructed example, a box contains ten labelled objects, four red and six blue. Select five without replacement. There are $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$ possible samples. The probabilities for $K=0,1,2,3,4$ areIn particular, $\mathbb{P}(K=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)/252=120/252=10/21$. Five red objects are impossible because the box only contains four. The expected value is $r(c/n)=2$, and the variance is $2/3$.

The count law also answers a detection question: what is the chance that five draws find at least one red object? By the complement rule of probability, it is $1-\mathbb{P}(K=0)=41/42$, approximately $0.9762$. This can model inspecting five distinct objects from a known finite collection, with a mark indicating the property we want to detect. If repeated inspections can select the same object again, the replacement model instead gives $1-0.6^5=0.92224$. Equal expected values therefore do not imply equal chances of detecting a marked object.

The corresponding binomial distribution counts red draws when each selected object is put back before the next independent draw: $K_{\mathrm{rep}}\sim \mathrm{Bin}(5,0.4)$. It has the same expected value $2$, but variance $5(0.4)(0.6)=6/5$, and it permits five red draws. Without replacement, a red first draw leaves red probability $3/9$ for the next draw; a blue first draw leaves $4/9$. This statistical dependence pulls the total toward the fixed composition of the population, producing the finite population correction.

The useful modelling question is whether previously selected objects remain available. A fixed-size sample of distinct items calls for a hypergeometric distribution; fresh independent trials with constant success probability call for a binomial distribution. When the population grows with $c/n\to p$ while $r$ stays fixed, the hypergeometric masses tend to the corresponding binomial distribution masses: removing finitely many objects has an increasingly small effect on the next draw's probability.

For uniform sampling without replacement from a population of $n>1$ objects, the finite population correction accounts for the reduced spread caused by drawing distinct objects. For the marked-object count $K$ in a hypergeometric distribution, put $p=c/n$. ThenRelative to the corresponding binomial distribution, the variance is multiplied by $f^2=(n-r)/(n-1)$ and the standard deviation by $f$. The square-root convention for the correction is used in Philip Stark's discussion of sample sums.^[ref]

The factor has a direct interpretation. For $r=1$, replacement makes no difference and $f=1$. For $r=n$, the entire population is observed and the marked count is exactly $c$, so $f=0$. In the ten-object, five-draw example, $f^2=5/9$ and $(6/5)(5/9)=2/3$. The exceptional population size $n=1$ is deterministic and should be handled directly rather than inserting it into a zero denominator.

A short explanation uses an indicator random variable $I_j$ for a marked object at draw $j$. Each has expected value $p$ and variance $p(1-p)$. For distinct draw positions $i,j$,Since $K={\sum }_{j=1}^r{I}_j$, the variance of a sum includes $r$ individual terms and $r(r-1)$ ordered covariance terms, giving the displayed correction. Linearity of expectation still gives $rp$ despite this dependence.

For arbitrary fixed real population values, use $N>1$ for population size and $1\le n\le N$ for sample size. Write the finite population mean $\bar{Y}=N^{-1}{\sum }_i{y}_i$, the finite population variance $S^2=(N-1{)}^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$ and ${\sigma }_N^2=N^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$. Under simple random sampling without replacement, the sample mean satisfiesThus $1-n/N$ is the variance factor relative to $S^2/n$, while $(N-n)/(N-1)$ is the exact variance ratio to independent uniform sampling with replacement, whose sample mean has variance ${\sigma }_N^2/n$. The standard error uses the square root of the corresponding factor. These formulas agree because ${\sigma }_N^2=(N-1)S^2/N$; the denominator convention must be specified. The sample mean under simple random sampling gives a derivation and an exact six-value example.

Suppose a finite population is split into $m$ categories with fixed sizes $c_1,\dots ,c_m$ summing to $n$. A uniform size-$r$ sample drawn by sampling without replacement has category counts $K_1,\dots ,K_m$ with the displayed joint probability mass function, for integers $0\le k_i\le c_i$ summing to $r$. It extends the two-category hypergeometric distribution; replacing the sampled objects instead gives a multinomial distribution.

For a box with category sizes $(5,3,2)$ and a sample of four distinct objects, the count pattern $(2,1,1)$ has probabilityPut $p_i=c_i/n$. When $n>1$, its covariance matrix is the corresponding multinomial distribution covariance matrix multiplied by the squared finite population correction:This connects the fixed-total dependence between categories with the further reduction from sampling without replacement. X. G. Duan gives a symmetry-based explanation of this shared covariance structure.^[ref]