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Hypergeometric Distribution Calculator

Math Probability • Discrete Probability Distributions

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Hypergeometric Distribution Calculator – P(X=k) Without Replacement (Free)

Compute hypergeometric probabilities for sampling without replacement: \(P(X=k)=\dfrac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}\). Visualize the PMF and simulate an actual draw from the urn.

Tip: Press Play after calculating to animate drawing \(n\) items without replacement and compare the simulated \(k\) to the PMF.

Model inputs
Population and draw

Accepted expressions: 1e-3, pi, e, sqrt(2), sin(), cos(), tan(), ln(), log(), abs(). Use * for multiplication.

Output & checks

Valid range: \(0 \le K \le N\), \(0 \le n \le N\), and feasible \(k\) must satisfy \(\max(0, n-(N-K)) \le k \le \min(n,K)\).

Visualization & animation
Drag on the PMF chart to pan. Use mouse wheel / trackpad to zoom. Tick labels stay inside the frame.
Ready
Interactive view — urn draw animation + PMF (pan/zoom)

Top: urn population (success/failure) and animated sample draws. Bottom: PMF bars \(P(X=x)\) with highlight at your \(k\) and the simulated \(k\).

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Frequently Asked Questions

When should I use the hypergeometric distribution?

Use it when sampling without replacement from a finite population that contains K successes and N−K failures.

What is the feasible range for k?

k must satisfy max(0, n-(N-K)) ≤ k ≤ min(n, K). Values outside this range have probability 0.

How is this different from a binomial distribution?

Binomial trials are independent with constant p; hypergeometric draws are dependent because the population changes after each draw.

Why does the variance include a correction factor?

Without replacement, draws are dependent. The finite population correction (N−n)/(N−1) reduces variance compared to an independent model.