Hypergeometric Distribution

Study guide: Exam P-1

The Hypergeometric Distribution models the number of successes in a sample of size n drawn without replacement from a finite population of N items, K of which are successes. Unlike the Binomial, trials are not independent.

P(X = k) = \frac{\dbinom{K}{k}\dbinom{N-K}{n-k}}{\dbinom{N}{n}}

k = \max(0,\,n+K-N),\ldots,\min(n,K)

\text{where } N = \text{population size},\; K = \text{successes in population},\; n = \text{sample size}

E[X] = \frac{nK}{N}

\text{Var}(X) = \frac{nK(N-K)(N-n)}{N^2(N-1)}

Drawing Defective Items A box contains 10 insurance policies: 4 with errors and 6 without. An auditor selects 3 at random without replacement. Find the probability exactly 2 have errors.

Answer N=10, K=4, n=3, k=2: P(X=2) = \frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} = \frac{6 \cdot 6}{120} = \frac{36}{120} = 0.30