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Hypergeometric Distribution
Compute the probability of k successes in n draws without replacement, plus mean and variance.
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P(X=k)=C(K,k)C(N−K,n−k)/C(N,n)
Population N
Successes K
Draws n
Draw successes k
Hypergeometric calculation
P(X=k) =
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📖 Tutorial | Hypergeometric Distribution
1. Definition
Hypergeometric distribution
draws n items without replacement from a population of N containing K successes; P(X=k)=C(K,k)C(N−K,n−k)/C(N,n).
2. Symbols
Symbol
Meaning
N
population size
K
successes in population
n
number drawn (without replacement)
k
successes drawn
3. How it works
Drawing without replacement, so not independent;
Probability is a ratio of combinations;
Mean E(X)=nK/N;
Variance has finite-population correction (N−n)/(N−1).
4. Steps
Enter N, K, n, k;
Click Calculate for P(X=k);
Read the mean and variance;
Check k≤min(K,n).
5. Example
Example:
a 52-card deck has 13 diamonds; draw 5, exactly 2 diamonds. P=C(13,2)C(39,3)/C(52,5)≈0.2743.
6. Pitfalls
Without replacement differs from binomial (with replacement);
k cannot exceed K or n;
It approximates binomial when N is large.
❓ FAQ | Hypergeometric Distribution
Difference from the binomial?
Hypergeometric: without replacement, not independent; binomial: with replacement.
Why is the mean nK/N?
Drawing n from success proportion K/N gives mean n×K/N.
When is it approx. binomial?
When N is much larger than n (e.g. n/N<0.05).
Which k are valid?
Integers from max(0,n−(N−K)) to min(K,n).
Where is it used?
Quality sampling, poker hands, estimation without replacement.