Suppose a finite population of size \(N\) contains \(M\) items of type "success" and \(N - M\) of type "failure". A random sample of size \(n\) is drawn without replacement. Let \(X\) be the number of successes in the sample. Then \(X\) follows a Hypergeometric distribution with parameters \(N, M, n\):
\[ P(X = x) \;=\; \dfrac{\binom{M}{x}\binom{N - M}{n - x}}{\binom{N}{n}}, \]for \(\max(0,\, n - (N-M)) \le x \le \min(n, M)\).
Compare with Binomial: Binomial samples with replacement (independent trials); Hypergeometric samples without replacement (dependent trials).
Validity: \(\sum_{x} \binom{M}{x}\binom{N-M}{n-x} = \binom{N}{n}\) (Vandermonde identity), so the PMF sums to 1.
where \(p = M/N\) and \(q = 1 - p\).
The factor \(\dfrac{N-n}{N-1}\) is called the finite population correction (FPC). It reduces the variance below the binomial value when sampling without replacement.
Sketch of mean derivation: Express \(X = \sum_{i=1}^{n} I_i\) where \(I_i = 1\) if the \(i\)-th item drawn is a success. By symmetry \(E(I_i) = M/N\), so \(E(X) = n M/N\).
A box contains 20 items of which 5 are defective. A random sample of 4 is drawn without replacement. Find the probability that exactly 2 are defective.
\(N = 20, M = 5, n = 4, x = 2\).
\(P(X = 2) = \dfrac{\binom{5}{2}\binom{15}{2}}{\binom{20}{4}} = \dfrac{10 \times 105}{4845} = \dfrac{1050}{4845} = 0.2167\).
From a deck of 52 cards, 5 cards are drawn without replacement. Find the expected number of aces and the variance.
\(N = 52, M = 4, n = 5\). Mean = \(5 \times 4/52 = 20/52 = 0.385\).
\(p = 4/52, q = 48/52\). Variance = \(5(0.077)(0.923)(47/51) \approx 0.327\).
Useful for computing successive probabilities without repeated factorial evaluation.
For \(N = 20, M = 5, n = 4\): \(P(X = 0) = \binom{15}{4}/\binom{20}{4} = 1365/4845 = 0.2817\).
\(P(X = 1) = \dfrac{(5)(4)}{(1)(12)} \cdot 0.2817 = \dfrac{20}{12} \times 0.2817 = 0.4696\).
\(P(X = 2) = \dfrac{(4)(3)}{(2)(13)} \times 0.4696 = \dfrac{12}{26} \times 0.4696 = 0.2167\) ✓ (matches direct).
Continuing: \(P(X = 3) = \dfrac{(3)(2)}{(3)(14)} \times 0.2167 = \dfrac{6}{42} \times 0.2167 = 0.0310\).
\(P(X = 4) = \dfrac{(2)(1)}{(4)(15)} \times 0.0310 = \dfrac{2}{60} \times 0.0310 = 0.00103\).
Sum = 0.2817 + 0.4696 + 0.2167 + 0.0310 + 0.00103 ≈ 1.000 ✓.
If \(N \to \infty\) and \(M \to \infty\) such that \(M/N \to p\) (a constant), with \(n\) fixed, then
\[ \dfrac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}} \;\to\; \binom{n}{x} p^x q^{n-x}. \]So the Hypergeometric approaches the Binomial when the population is much larger than the sample.
Practical rule: If \(n/N < 0.05\) (sample less than 5% of population), Binomial is a good approximation to Hypergeometric.
The variance reflects this: \(\dfrac{N-n}{N-1} \to 1\) as \(N \to \infty\), so Hypergeometric variance \(\to\) Binomial variance \(npq\).