Hypergeometric models

What you will learn

Derive a hypergeometric probability and check its feasible support.

Prerequisites: Combinations versus permutations · Runouts and outs

Choose successes and failures separately

A population has N distinct items, K marked successes and N−K failures. In an unordered uniform draw of n items without replacement, exactly k successes can be chosen in C(K,k) ways and the failures in C(N−K,n−k) ways. Multiply those independent choices of sets.

Divide by C(N,n), the number of all n-item subsets. The resulting hypergeometric probability is C(K,k)C(N−K,n−k)/C(N,n). The selections inside this counting argument are separate combinatorial choices; the sequential draws themselves are not independent.

Check feasibility before arithmetic

A success count must satisfy max(0,n−(N−K)) ≤ k ≤ min(K,n). For example, drawing four items from a five-item population with only two failures guarantees at least two successes. An impossible count has probability zero.

Summing probabilities over all feasible counts should give one. This is a useful check for omissions, but it does not establish that the real population was sampled uniformly or that success labels were correctly defined.

Common mistakes

• Using a binomial model that assumes replacement or independent trials.
• Choosing n failures instead of n−k failures.
• Computing factorials for impossible counts instead of checking support.

Keep these ideas

• Count favorable success and failure subsets.
• Use the same unordered convention for the denominator.
• Check support and the total probability mass.