Hypergeometric Distribution Calculator

Calculate exact hypergeometric probabilities for sampling without replacement, with PMF, CDF, distribution tables, and binomial approximation comparison.

About the Hypergeometric Distribution Calculator

The hypergeometric distribution calculator computes exact probabilities for sampling without replacement from a finite population. Each draw changes the remaining pool, so the probability of the next success is not constant the way it is in the binomial case.

Use it when the population is finite and items are not returned after each draw: inspection lots, card hands, grab samples, or any fixed-size batch where the composition shifts after every selection.

Enter the population size N, number of success states K, draw size n, and target successes k to get point probabilities, cumulative probabilities, and a full distribution table.

Why Use This Hypergeometric Distribution Calculator?

This calculator saves time whenever the population is fixed and the sampling is without replacement. That is the setting where binomial shortcuts can drift away from the true answer.

The distribution table is useful for spotting where the probability mass sits, and the binomial comparison shows whether the simpler approximation is close enough for your use case.

How to Use This Calculator

  1. Enter the total population size (N) — all items in the group.
  2. Enter the number of success states (K) — items with the desired property.
  3. Enter the draw size (n) — how many items you draw without replacement.
  4. Enter the observed successes (k) for point probability.
  5. Set range bounds x₁ and x₂ for interval probability calculations.
  6. Review the distribution chart, full table, and binomial comparison.
  7. Use presets for quick exploration of common scenarios.

Formula

P(X = k) = C(K,k) × C(N−K, n−k) / C(N,n), where C(a,b) = a! / (b!(a−b)!). Mean = nK/N. Variance = nK(N−K)(N−n) / (N²(N−1)).

Example Calculation

Result: P(X = 2) ≈ 0.2743 (27.43%)

Drawing 5 cards from a standard 52-card deck, the probability of getting exactly 2 hearts (13 in the deck) is C(13,2)×C(39,3) / C(52,5) ≈ 0.2743.

Tips & Best Practices

Why The Probability Changes

With replacement, each draw resets the population. Without replacement, every success or failure changes what remains. That is why the hypergeometric distribution is exact for finite sampling and the binomial is only an approximation.

Where It Shows Up

The same math appears in quality control, deck-of-cards calculations, sample audits, and exact tests on contingency tables. Any time you are drawing a fixed number of items from a known finite set, the hypergeometric model is the natural fit.

Reading The Approximation

The binomial comparison is a quick way to judge whether replacement assumptions are close enough. When the sample is small relative to the population, the two distributions are often close; when the sample is large, the difference becomes more important.

Sources & Methodology

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Frequently Asked Questions

When should I use hypergeometric vs. binomial?

Use hypergeometric for sampling without replacement from a finite population (card draws, quality inspection). Use binomial when draws are independent (coin flips, each trial identical).

What does "without replacement" mean?

Each item drawn is not returned to the pool, so the composition changes. Drawing 2 hearts makes the remaining deck have 11 hearts in 50 cards, not 13 in 52.

Can the hypergeometric distribution have k > K or k > n?

No. The number of successes k must be between max(0, n+K−N) and min(n, K). Values outside this range have probability zero.

How is this related to Fisher's exact test?

Fisher's exact test calculates the probability of observing a 2×2 contingency table (or more extreme) using the hypergeometric distribution. It's the gold standard for small-sample tests of independence.

What is acceptance sampling?

In quality control, you draw a sample of n items from a lot of N. If more than c defectives are found, the lot is rejected. The hypergeometric distribution gives exact acceptance/rejection probabilities.

Why does the binomial approximation column show differences?

The binomial assumes constant probability p = K/N for each draw. Without replacement, the actual probability changes with each draw. The difference shrinks as N grows relative to n.

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