Hypergeometric Distribution Calculator
Find the probability of drawing k successes in a sample of n taken without replacement from a population of N that contains K successes — exactly k, at most, at least, or a range — with the mean, variance, a full table, a bar chart and a comparison with sampling with replacement. Free, no sign-up.
Adds P(k ≤ X ≤ this value) to the results.
Results
| Probability | Meaning | Value |
|---|---|---|
| P(X = 2) | exactly 2 | 0.0399 |
| P(X ≤ 2) | at most 2 | 0.9982 |
| P(X < 2) | fewer than 2 | 0.9583 |
| P(X ≥ 2) | at least 2 | 0.0417 |
| P(X > 2) | more than 2 | 0.0018 |
k can only take values from 0 to 4 with these inputs.
0.3846
Mean (nK/N)
0.3272
Variance
0.572
Standard deviation
Drawn with replacement instead, the count would be binomial with n = 5 and p = K/N = 0.0769, giving P(X = 2) = 0.0465 against 0.0399 here. The finite population correction (N − n)/(N − 1) = 0.9216 is why the variance here is smaller.
Probability distribution
| k | P(X = k) | P(X ≤ k) | P(X ≥ k) |
|---|---|---|---|
| 0 | 0.6588 | 0.6588 | 1.0000 |
| 1 | 0.2995 | 0.9583 | 0.3412 |
| 2 | 0.0399 | 0.9982 | 0.0417 |
| 3 | 0.0017 | 1.0000 | 0.0018 |
| 4 | 1.85 × 10−5 | 1.0000 | 1.85 × 10−5 |
How to Use the Hypergeometric Distribution Calculator
- 1. Describe the population. N is the number of items in total and K the number of them that count as successes — 52 cards with 4 aces, a lot of 50 parts with 5 defectives.
- 2. Enter the sample size. n is how many items are drawn, without putting any back.
- 3. Enter k, and optionally a second value. k is the number of successes in the sample you are asking about; a second value adds the probability of a whole range.
- 4. Read the results. Every probability for k, the possible range of k, the mean, variance and standard deviation, the binomial answer for comparison, and a full table and bar chart.
What the Hypergeometric Distribution Models
The hypergeometric distribution counts successes in a sample drawn without replacement from a finite population: aces in a poker hand, defective parts in an inspection sample, winning numbers on a lottery ticket. Because nothing is put back, every draw changes the odds for the next one — exactly what the binomial distribution, with its constant p, cannot handle.
The probability of exactly k successes is P(X = k) = C(K, k) × C(N − K, n − k) / C(N, n): the ways to choose k of the K successes and n − k of the N − K failures, divided by the ways to choose any n items. The mean is nK/N, as for a binomial with p = K/N, but the variance is multiplied by the finite population correction: nK/N × (1 − K/N) × (N − n)/(N − 1).
| Feature | Hypergeometric | Binomial |
|---|---|---|
| Sampling | Without replacement | With replacement, or from an unlimited population |
| Chance of success | Changes after every draw | Constant p |
| Parameters | N, K, n | n, p |
| Variance | np(1 − p) × (N − n)/(N − 1), with p = K/N | np(1 − p) |
| Use it when | The sample is a sizeable share of the population | The sample is under about 5% of the population |
When the sample is only a small fraction of the population — under 5% is the usual rule — the correction is close to 1, the two distributions agree closely, and the binomial is a convenient approximation.
Worked Examples: Cards, Inspections and Lotteries
Cards. A 5-card hand from a 52-card deck holding 4 aces has N = 52, K = 4, n = 5. The chance of exactly 2 aces is C(4, 2) × C(48, 3) / C(52, 5) = 6 × 17,296 / 2,598,960 = 0.0399, and of at least one ace 0.3412. If each card were replaced and the deck reshuffled, the binomial would give 0.0465 for exactly two: drawing without replacement makes a second ace less likely, because the first one is no longer in the deck.
Quality control. A lot of 50 parts contains 5 defectives, and an inspector tests 10 at random. The chance of finding at least one defective is 0.6894 — so a lot this bad still passes an inspection that demands zero defectives about 31% of the time.
Lottery. In a 6/49 lottery, 6 of the 49 numbers win and you pick 6. Matching exactly 3 has probability 0.0177, about 1 in 57; matching all six has probability 7.15 × 10−8, or 1 in 13,983,816.
Frequently Asked Questions
- When should I use the hypergeometric distribution instead of the binomial?
- Whenever you sample without replacement from a finite population and the sample is more than a small fraction of it — roughly 5% or more. Each draw then changes the chance of success on the next, and only the hypergeometric accounts for that. For a small sample from a huge population the two give nearly the same answer.
- What is the formula for the hypergeometric distribution?
- P(X = k) = C(K, k) × C(N − K, n − k) / C(N, n), where N is the population size, K the number of successes in it, n the sample size and k the number of successes in the sample. C(a, b) is the number of ways to choose b items from a.
- What is the probability of getting 2 aces in a 5-card hand?
- Exactly 2 aces: 0.0399, about 1 in 25. At least one ace: 0.3412. Both come from N = 52 cards, K = 4 aces and n = 5 cards drawn without replacement.
- What are the mean and variance of a hypergeometric distribution?
- The mean is nK/N and the variance nK/N × (1 − K/N) × (N − n)/(N − 1). For a 5-card hand the expected number of aces is 5 × 4/52 = 0.3846, with a variance of 0.3272.
- Why can k only take some values?
- A sample cannot contain more successes than the population holds, or more failures. So k runs from max(0, n − (N − K)) to min(n, K): in a 5-card hand you can hold at most 4 aces, and if you drew 50 cards you would hold at least 2.
- How large can the population be?
- Up to 100,000. The cumulative probabilities are exact sums over every possible value of k, which stay instant at that size. For a larger population and a small sample, the binomial with p = K/N is an excellent approximation.