Hypergeometric Distribution Calculator

The hypergeometric distribution models the probability of obtaining exactly $k$ successes in $n$ draws, without replacement, from a finite population of size $N$ that contains exactly $K$ total successes. The core probability mass function is calculated using combinations: P(X = k) = [C(K, k) * C(N - K, n - k)] / C(N, n), where C(a, b) represents the binomial coefficient a-choose-b. Additionally, the calculator derives the expected mean using the formula Mean = (n * K) / N, and computes the variance by accounting for the finite population correction factor: Variance = [n * K * (N - K) * (N - n)] / [N^2 * (N - 1)].

Imagine a small business quality audit where a shipment of 50 electrical components contains 10 defective items. An inspector randomly selects a sample of 5 components without replacement. Let us calculate the probability of finding exactly 1 defective component in this sample. Here, the total population (N) is 50, total successes or defectives (K) is 10, sample size (n) is 5, and our target number of successes (k) is 1. Using the formula, we find the combinations: C(10, 1) = 10, C(40, 4) = 91,390, and C(50, 5) = 2,118,760. Multiplying the numerator combinations and dividing by the total population combination gives a probability of approximately 0.4314, or 43.14%. Furthermore, the expected mean number of defective parts in any 5-part sample is (5 * 10) / 50 = 1.0, with a variance of approximately 0.7347.

Always verify whether your sampling process involves replacement or not; if items are replaced after each draw, you should use the binomial distribution instead of the hypergeometric model. Ensure your total population size (N) is accurate and strictly greater than your sample size (n). When dealing with large populations exceeding several thousand items, the hypergeometric distribution converges closely with the binomial distribution, though hypergeometric remains the gold standard for small, finite batches.