Skip to main content

Hypergeometric distribution; binomial approximation

Short answer

Hypergeometric distribution models successes in sampling without replacement from a finite population. Binomial approximation holds when population N is large, sample size n/N ≤ 0.05, and p=K/N not extreme; then probabilities are approximately binomial with p=K/N.

Define hypergeometric probability distribution with an example. Describe the conditions for the binomial approximation to hypergeometric distribution?

Hypergeometric distribution

Suppose a finite population of N items has K "successes" (and N − K "failures"). If n items are selected at random without replacement, the number of successes X has the hypergeometric distribution:

P(X = x) = C(K, x) · C(N − K, n − x) / C(N, n), max(0, n − N + K) ≤ x ≤ min(n, K)

Mean = nK/N; variance = n (K/N)(1 − K/N)(N − n)/(N − 1).

Example: A box has 10 bulbs, 4 defective. Three are drawn without replacement. The probability that exactly 1 is defective is C(4,1) C(6,2)/C(10,3) = 4 × 15 / 120 = 0.5.

Binomial approximation to the hypergeometric

Draws without replacement become almost independent when the population is very large compared with the sample, so the hypergeometric can be replaced by B(n, p) with p = K/N:

C(K, x) C(N − K, n − x) / C(N, n) ≈ C(n, x) pˣ (1 − p)ⁿ⁻ˣ

Conditions:

  1. The population size N is large.
  2. The sample size is small compared with N, normally n/N ≤ 0.05 (some books say 0.1), so the finite population factor (N − n)/(N − 1) ≈ 1.
  3. The proportion of successes p = K/N stays almost constant during sampling (K is also large, and p is not extremely close to 0 or 1).

Then mean nK/N ≈ np and variance ≈ npq.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 2 questions →