In sampling without replacement, what distribution describes the number of successes in a fixed-size sample?

• A. Hypergeometric distribution; P(X=k) = [C(K, k) C(N − K, n − k)] / C(N, n)
• B. Binomial distribution; P(X=k) = C(n, k) p^k (1−p)^{n−k}
• C. Poisson distribution; P(X=k) = e^{−λ} λ^k / k!
• D. Geometric distribution; P(X=k) = (1−p)^{k−1} p

Answer: (A)

When you draw a fixed number of items from a finite population that has a known number of successes, the number of successes in your sample follows a hypergeometric distribution. The reason is that each draw changes the composition of the population, so the trials are not independent.

The probability of getting exactly k successes in the sample is P(X = k) = [C(K, k) C(N − K, n − k)] / C(N, n). This counts the ways to choose k successes from the K available and n − k failures from the N − K non-successes, divided by all possible ways to choose n items from N.

This differs from the binomial distribution, which assumes independent trials with a constant success probability on each draw—an assumption that fails here because removing items without replacement updates those probabilities. The Poisson and geometric distributions describe other scenarios (rare events in a large population, or the number of trials until the first success in independent trials, respectively) and don’t match the fixed-size, without-replacement setup.