Hypergeometric Distribution Table

Count successes without replacement

N is the population size, K the number of success items, n the number drawn uniformly without replacement, and X the success count. Its support is max(0, n − (N − K)) through min(n,K), inclusive.

$P(X=k)=\frac{\binom Kk\binom{N-K}{n-k}}{\binom Nn}.$

The expected count is nK/N. For N > 1, the variance is n(K/N)(1 − K/N)(N − n)/(N − 1). Draws are dependent because the population composition changes. A binomial model uses a different sampling mechanism.

Cells show exact probabilities P(X = k), rounded to four decimals. Impossible outcomes appear as 0.0000, but other 0.0000 cells can be positive probabilities rounded down.

Exact probabilities: N = 20, n = 5

k	K = 2	K = 3	K = 4	K = 5	K = 6	K = 7	K = 8	K = 10
0	0.5526	0.3991	0.2817	0.1937	0.1291	0.0830	0.0511	0.0163
1	0.3947	0.4605	0.4696	0.4402	0.3874	0.3228	0.2554	0.1354
2	0.0526	0.1316	0.2167	0.2935	0.3522	0.3874	0.3973	0.3483
3	0.0000	0.0088	0.0310	0.0677	0.1174	0.1761	0.2384	0.3483
4	0.0000	0.0000	0.0010	0.0048	0.0135	0.0293	0.0542	0.1354
5	0.0000	0.0000	0.0000	0.0001	0.0004	0.0014	0.0036	0.0163

Exact probabilities: N = 50, n = 10

k	K = 5	K = 10	K = 15	K = 20	K = 25
0	0.3106	0.0825	0.0179	0.0029	0.0003
1	0.4313	0.2662	0.1031	0.0279	0.0050
2	0.2098	0.3369	0.2406	0.1083	0.0316
3	0.0442	0.2178	0.2979	0.2259	0.1076
4	0.0040	0.0785	0.2157	0.2801	0.2181
5	0.0001	0.0161	0.0949	0.2151	0.2748
6	0.0000	0.0019	0.0255	0.1034	0.2181
7	0.0000	0.0001	0.0041	0.0306	0.1076
8	0.0000	0.0000	0.0004	0.0053	0.0316
9	0.0000	0.0000	0.0000	0.0005	0.0050
10	0.0000	0.0000	0.0000	0.0000	0.0003

Exact probabilities: N = 100, n = 10

k	K = 5	K = 10	K = 15	K = 20
0	0.5838	0.3305	0.1808	0.0951
1	0.3394	0.4080	0.3568	0.2679
2	0.0702	0.2015	0.2919	0.3182
3	0.0064	0.0518	0.1297	0.2092
4	0.0003	0.0076	0.0345	0.0841
5	0.0000	0.0006	0.0057	0.0215
6	0.0000	0.0000	0.0006	0.0035
7	0.0000	0.0000	0.0000	0.0004
8	0.0000	0.0000	0.0000	0.0000
9	0.0000	0.0000	0.0000	0.0000
10	0.0000	0.0000	0.0000	0.0000

Exact probabilities: N = 100, n = 20

k	K = 5	K = 10	K = 15	K = 20
0	0.3193	0.0951	0.0262	0.0066
1	0.4201	0.2679	0.1191	0.0433
2	0.2073	0.3182	0.2363	0.1259
3	0.0478	0.2092	0.2711	0.2159
4	0.0051	0.0841	0.2004	0.2437
5	0.0002	0.0215	0.1008	0.1920
6	0.0000	0.0035	0.0355	0.1091
7	0.0000	0.0004	0.0089	0.0456
8	0.0000	0.0000	0.0016	0.0142
9	0.0000	0.0000	0.0002	0.0033
10	0.0000	0.0000	0.0000	0.0006
11	0.0000	0.0000	0.0000	0.0001
12	0.0000	0.0000	0.0000	0.0000
13	0.0000	0.0000	0.0000	0.0000
14	0.0000	0.0000	0.0000	0.0000
15	0.0000	0.0000	0.0000	0.0000
16	0.0000	0.0000	0.0000	0.0000
17	0.0000	0.0000	0.0000	0.0000
18	0.0000	0.0000	0.0000	0.0000
19	0.0000	0.0000	0.0000	0.0000
20	0.0000	0.0000	0.0000	0.0000

Worked example

Select two items uniformly from a hypothetical box of ten containing three defective items. Exactly one defect has probability C(3,1)C(7,1)/C(10,2) = 21/45 ≈ 0.4667. The binomial value with p = 0.3 would be 0.42 and is not exact for this design.

For a cumulative probability, sum exact cells across the desired support. These finite-support tables cover the whole support in each column; rounded sums may differ slightly from 1. Do not infer an upper tail from a truncated set of rows without accounting for omitted outcomes.

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Study sampling methods and discrete distributions. SciPy specifies the model, support and parameter convention.