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Hypergeometric Distribution Calculator

The probability of drawing a given number of successes when sampling without replacement — cards from a deck, defectives from a batch, an audit sample from a ledger.

When you sample without replacement the draws are not independent: taking one red ball changes the odds for the next.

Results update as you type
Results
P(X = k)
29.947364%
P(X ≤ k)
P(X ≥ k)
P(none at all)
Expected successes
Standard deviation
Binomial approximation
Reviewed September 2026. Pure mathematics: the result does not depend on where you are. Terminology follows US usage (math, parentheses, decimal point).
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About hypergeometric distribution

How the hypergeometric distribution calculator works

When you sample without replacement the draws are not independent: taking one red ball changes the odds for the next. The hypergeometric distribution handles that exactly. With K successes in a population of N and a sample of n, the chance of exactly k successes is C(K, k) C(N − K, n − k) ÷ C(N, n).

If the population is very large relative to the sample, the binomial distribution is a good approximation — the calculator shows both so you can see when it stops being one.

Formula: P(X = k) = C(K, k) C(N − K, n − k) / C(N, n)

Worked examples

InputsP(X = k)Note
Exactly one ace in a 5-card hand29.947364%about 29.9%
No aces in a 5-card hand65.8842%about 65.9%
2 defectives in a sample of 10 from 100 with 5 bad7.021881%about 7%

Frequently asked questions

When do I use the hypergeometric rather than the binomial?

Whenever you sample without replacement from a finite population. If the sample is under about 5% of the population the two agree closely.

What is the chance of at least one ace in a five-card hand?

About 34.1% — the complement of the 65.9% chance of none.

Why are the draws not independent?

Because removing a card changes what is left. After drawing an ace, only three remain in 51 cards.

Where is this used in practice?

Acceptance sampling and audit: a sample of 10 from a batch of 100 with 5 defectives finds at least one about 42% of the time — which is why small samples miss small defect rates.

What is the expected number of successes?

n × K / N, exactly the same as the binomial. Only the spread differs, by the finite-population correction.

Where these figures come from

Last checked: September 2026. Formulas are fixed by mathematics and do not change with tax years or regulations.