Hypergeometric Distribution Calculator

The hypergeometric distribution gives the probability of drawing exactly k successes in n draws made without replacement from a finite population of size N that contains K successes. Its probability mass function is P(X = k) = C(K,k) × C(N−K,n−k) / C(N,n), where C(a,b) is the binomial coefficient. Use it for card hands, lottery draws, and quality-control sampling — anywhere each draw changes the odds because items are not put back.

Treat hearts as successes: N = 52 cards, K = 13 hearts, n = 5 drawn, k = 2 observed. P(X = 2) = C(13,2) × C(39,3) / C(52,5) = 9139/33320 ≈ 0.2742797118847539, about 27.427971%. The mean is nK/N = 1.25 hearts and P(X ≤ 2) = 0.9072328931572629.

A 6/49 lottery draws n = 6 balls from N = 49 with K = 6 winning balls. Matching all 6 (k = 6) has probability P(X = 6) = 1 / C(49,6) = 1/13983816 ≈ 7.151123842018516e-08, exactly 1 in 13,983,816. The expected number of matches per ticket is nK/N = 36/49 ≈ 0.7346938775510204.

When the sample is a tiny fraction of the population — the usual rule is n/N ≤ 5% — sampling without replacement barely changes the odds, so the binomial with p = K/N is a fine approximation. Example: N = 100000, K = 20000, n = 20, k = 5. Hypergeometric P(X = 5) = 0.17457479735358877, binomial(20, 0.2) gives 0.17455952155688043, a relative difference of only 0.00875%. This calculator shows the comparison automatically when n/N ≤ 0.05.

QC inspectors sample n items from a batch of N and count defectives. Example: a batch of N = 500 with K = 25 defectives, sample n = 30. The probability of finding exactly k = 1 defective is P(X = 1) ≈ 0.3441179773040583 (34.411798%). The expected count is nK/N = 1.5 with standard deviation ≈ 1.1585268096757515. Acceptance-sampling plans use P(X ≤ c) — computed in the results table above — to decide whether to accept the batch.

Mean μ = nK/N. Variance σ² = n × (K/N) × ((N−K)/N) × ((N−n)/(N−1)), where the last factor (N−n)/(N−1) is the finite population correction. Example: N = 20, K = 8, n = 6 gives μ = 2.4, σ² = 1.0610526315789475, σ ≈ 1.0300740903347425. As N grows large relative to n, the correction approaches 1 and the variance approaches the binomial value np(1−p).

k must satisfy max(0, n − (N−K)) ≤ k ≤ min(n, K). You cannot draw more successes than exist (k ≤ K) or than you draw (k ≤ n), and if the draw is large relative to the failures you are forced to include successes: with N = 52, K = 13 hearts, n = 5, k ranges from 0 to 5. Values outside this range have probability exactly 0, which the calculator reports instead of erroring.

Because each draw changes the composition of what remains. After drawing one heart from a 52-card deck, 12 hearts remain out of 51 cards — the probability shifts every draw. Ignoring this and using the binomial anyway overstates the variance: for the 5-card hearts example the true variance is 0.8639705882352942 (finite population correction 47/51), versus 0.9375 under the binomial. For large populations relative to the sample the difference vanishes.