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Hypergeometric Distribution



Key Terms
The Probabilistic Experiment Behind hypergeometric distribution
Notation Used
Parameters
Probability Mass Function (PMF) and Support (Range)
Cumulative Distribution Function (CDF)
Expected Value (Mean)
Variance and Standard Deviation
Mode and Median
Applications and Examples
Interactive Calculator
Hypergeometric at a Glance



Hypergeometric Distribution: Sampling Without Replacement


The hypergeometric distribution arises from sampling without replacement from a finite population containing two types of items, typically labeled success and failure. A fixed number of draws is made, and the random variable counts how many successes appear in the sample. Because items are not replaced, each draw changes the probabilities of subsequent draws, introducing dependence between trials — the key distinction from the binomial experiment.

Key Terms

Hypergeometric Distribution— successes in nn draws without replacement
Binomial Distribution— the analogous model for sampling with replacement
Expected Value— E[X]=nK/NE[X] = nK/N
Variance— Var⁡(X)\operatorname{Var}(X) with finite population correction

See All Probability Definitions →


The Probabilistic Experiment Behind hypergeometric distribution


The hypergeometric distribution counts the number of successes obtained when sampling without replacement from a finite population. The population contains a fixed number of successes and failures, and each draw permanently alters the composition of the population.

Unlike the binomial distribution, trials are not independent. The probability of success changes after each draw because items are not returned. The number of draws is fixed in advance, and the random variable counts how many successes appear in the sample.

This distribution captures situations where resources are limited or where selection without replacement is intrinsic to the experiment. It reflects dependence between outcomes — a key distinction from trial-based models.

Example:

Drawing 55 cards from a standard deck without replacement and counting how many are hearts. Each draw changes the probabilities for subsequent draws, because the deck composition changes.

Notation Used


X∼Hypergeometric(N,K,n)X \sim \text{Hypergeometric}(N, K, n) or X∼Hyp(N,K,n)X \sim \text{Hyp}(N, K, n) — distribution of the random variable.

Hypergeometric(N,K,n)\text{Hypergeometric}(N, K, n) — used to denote the distribution itself (not the random variable).

H(N,K,n)\text{H}(N, K, n) — occasionally used in compact form, especially in software or formulas.

P(X=k)=(Kk)(N−Kn−k)(Nn),for valid kP(X = k) = \frac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}}, \quad \text{for valid } k — probability mass function (PMF), where:

NN — total population size

KK — number of success states in the population

nn — number of draws (sample size) without replacement

kk — number of observed successes in the sample

max⁡(0,n−(N−K))≤k≤min⁡(n,K)\max(0, n - (N - K)) \leq k \leq \min(n, K) — valid range for kk

(ab)=a!b!(a−b)!\binom{a}{b} = \frac{a!}{b!(a-b)!} — binomial coefficient

Key properties:

E[X]=nKNE[X] = n \frac{K}{N} — expected value (mean)

Var(X)=nKNN−KNN−nN−1\text{Var}(X) = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1} — variance

Relationship to binomial distribution:

Hypergeometric(N,K,n)≈Binomial(n,p)\text{Hypergeometric}(N, K, n) \approx \text{Binomial}(n, p) where p=KNp = \frac{K}{N}, when NN is large relative to nn (sampling with replacement approximation)

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Parameters


𝑁𝑁: total population size

𝐾𝐾: number of successes in the population

𝑛𝑛: number of draws (without replacement), where 𝑛≤𝑁𝑛≤𝑁

The hypergeometric distribution models the number of successes in 𝑛𝑛 draws from a finite population of size 𝑁𝑁 that contains exactly 𝐾𝐾 successes, without replacement.

Unlike the binomial distribution, where each trial is independent, here each draw changes the probabilities — once an item is drawn, it doesn't go back. This dependency is what defines the distribution's behavior.

Probability Mass Function (PMF) and Support (Range)


The probability mass function (PMF) of a hypergeometric distribution is given by:

P(X=k)=(Kk)(N−Kn−k)(Nn),k=max⁡(0,n−N+K),…,min⁡(n,K)P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}, \quad k = \max(0, n-N+K), \ldots, \min(n, K)


where (ab)=a!b!(a−b)!\binom{a}{b} = \frac{a!}{b!(a-b)!} is the binomial coefficient.


Sampling Without Replacement: The hypergeometric distribution models the number of successes when drawing nn items without replacement from a finite population of size NN containing exactly KK success items.

Support (Range of the Random Variable):
* The random variable XX can take on values from max⁡(0,n−N+K)\max(0, n-N+K) to min⁡(n,K)\min(n, K).
* X=kX = k means exactly kk successes are drawn in the sample of size nn.
* The lower bound ensures we don't draw more failures than available: n−k≤N−Kn-k \leq N-K
* The upper bound ensures we don't draw more successes than available: k≤Kk \leq K and k≤nk \leq n
* The support is thus a finite set of non-negative integers.

Logic Behind the Formula:
* (Kk)\binom{K}{k}: The number of ways to choose kk successes from KK available successes
* (N−Kn−k)\binom{N-K}{n-k}: The number of ways to choose n−kn-k failures from N−KN-K available failures
* (Nn)\binom{N}{n}: The total number of ways to choose nn items from NN items
* The total probability sums to 1:

∑k=max⁡(0,n−N+K)min⁡(n,K)P(X=k)=∑k=max⁡(0,n−N+K)min⁡(n,K)(Kk)(N−Kn−k)(Nn)=1\sum_{k=\max(0,n-N+K)}^{\min(n,K)} P(X = k) = \sum_{k=\max(0,n-N+K)}^{\min(n,K)} \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}} = 1

* This follows from Vandermonde's identity.

Hypergeometric Distribution

Sampling without replacement from finite population

Explanation

The hypergeometric distribution models the number of successes when sampling without replacement from a finite population. The probability mass function is P(X=k)=(Kk)(N−Kn−k)(Nn)P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}, where NN is the population size, KK is the number of success states in the population, and nn is the number of draws. The expected value is E[X]=n⋅KNE[X] = n \cdot \frac{K}{N} and the variance is Var(X)=n⋅KN⋅(1−KN)⋅N−nN−1\text{Var}(X) = n \cdot \frac{K}{N} \cdot \left(1-\frac{K}{N}\right) \cdot \frac{N-n}{N-1}. Applications include finding defective items in a sample, analyzing card hands dealt without replacement, and quality control sampling.


Cumulative Distribution Function (CDF)


    The cumulative distribution function (CDF) of a hypergeometric distribution is given by:

    FX(k)=P(X≤k)=∑i=0k(Ki)(N−Kn−i)(Nn)F_X(k) = P(X \leq k) = \sum_{i=0}^{k} \frac{\binom{K}{i}\binom{N-K}{n-i}}{\binom{N}{n}}


    Where:
    NN = total population size
    KK = number of success states in the population
    nn = number of draws (sample size)
    kk = number of observed successes in the sample (where max⁡(0,n−N+K)≤k≤min⁡(n,K)\max(0, n-N+K) \leq k \leq \min(n, K))

    Intuition Behind the Formula


    Definition: The CDF gives the probability of observing kk or fewer successes when drawing nn items without replacement from a population of size NN containing KK success states.

    Summation of Probabilities:
    We sum the individual probabilities from the minimum possible value up to kk:

    P(X≤k)=P(X=0)+P(X=1)+P(X=2)+⋯+P(X=k)P(X \leq k) = P(X=0) + P(X=1) + P(X=2) + \cdots + P(X=k)


    Without Replacement Effect: Unlike the binomial distribution, the hypergeometric CDF accounts for sampling without replacement. Each draw changes the composition of the remaining population, creating dependency between draws.

    Boundary Conditions:
    The support is bounded by physical constraints:
  • max⁡(0,n−N+K)\max(0, n-N+K) (can't draw more failures than exist)
  • min⁡(n,K)\min(n, K) (can't draw more successes than exist or more items than drawn)

    • Complementary Probability:
    For "more than kk successes":

    P(X>k)=1−FX(k)P(X > k) = 1 - F_X(k)

Hypergeometric Distribution CDF

CDF for sampling without replacement

CDF Explanation

The hypergeometric CDF is F(k)=P(X≤k)=∑i=0k(Ki)(N−Kn−i)(Nn)F(k) = P(X \leq k) = \sum_{i=0}^{k} \frac{\binom{K}{i} \binom{N-K}{n-i}}{\binom{N}{n}} for max⁡(0,n−(N−K))≤k≤min⁡(n,K)\max(0, n-(N-K)) \leq k \leq \min(n, K). This represents the probability of drawing kk or fewer success items when sampling nn items without replacement from a population of size NN containing KK success items. The CDF is bounded by the minimum and maximum possible number of successes in the sample. Unlike the binomial CDF, the hypergeometric CDF accounts for the changing probability as items are drawn without replacement.

Expected Value (Mean)


As explained in the general case for calculating expected value, the expected value of a discrete random variable is computed as a weighted sum where each possible value is multiplied by its probability:

E[X]=∑xx⋅P(X=x)E[X] = \sum_{x} x \cdot P(X = x)


For the hypergeometric distribution, we apply this general formula to the specific probability mass function of this distribution.

Formula


E[X]=n⋅KNE[X] = n \cdot \frac{K}{N}


Where:
NN = total population size
KK = number of success states in the population
nn = number of draws (sample size)

Derivation and Intuition


The hypergeometric distribution describes sampling without replacement. Although the draws are not independent, the expected value has a remarkably simple form.

The proportion of successes in the population is KN\frac{K}{N}. When drawing nn items, each draw has the same marginal probability KN\frac{K}{N} of being a success (even though the draws are dependent).

By symmetry and linearity of expectation, the expected number of successes in nn draws is:

E[X]=n⋅KNE[X] = n \cdot \frac{K}{N}


The result E[X]=n⋅KNE[X] = n \cdot \frac{K}{N} captures the intuition that the expected proportion of successes in the sample matches the proportion in the population. If you sample nn items from a population where the success rate is KN\frac{K}{N}, you expect n⋅KNn \cdot \frac{K}{N} successes on average.

Example


Consider drawing 5 cards from a standard deck of 52 cards, counting the number of aces. Here N=52N = 52, K=4K = 4, and n=5n = 5:

E[X]=5⋅452=2052≈0.385E[X] = 5 \cdot \frac{4}{52} = \frac{20}{52} \approx 0.385


On average, you expect to draw about 0.385 aces in a 5-card hand, which reflects the 4-in-52 proportion of aces in the deck.

Variance and Standard Deviation


The variance of a discrete random variable measures how spread out the values are around the expected value. It is computed as:

Var(X)=E[(X−μ)2]=∑x(x−μ)2P(X=x)\mathrm{Var}(X) = \mathbb{E}[(X - \mu)^2] = \sum_{x} (x - \mu)^2 P(X = x)


Or using the shortcut formula:

Var(X)=E[X2]−μ2\mathrm{Var}(X) = \mathbb{E}[X^2] - \mu^2


For the hypergeometric distribution, we apply this formula to derive the variance.

Formula


Var(X)=n⋅KN⋅N−KN⋅N−nN−1\mathrm{Var}(X) = n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1}


Where:
NN = total population size
KK = number of success states in the population
nn = number of draws (sample size)

Derivation and Intuition


The derivation involves computing E[X2]\mathbb{E}[X^2] using indicator random variables and accounting for the dependency created by sampling without replacement.

We know from the expected value section that μ=n⋅KN\mu = n \cdot \frac{K}{N}.

The variance formula can be rewritten to show its relationship to the binomial variance:

Var(X)=n⋅KN⋅(1−KN)⋅N−nN−1\mathrm{Var}(X) = n \cdot \frac{K}{N} \cdot \left(1 - \frac{K}{N}\right) \cdot \frac{N-n}{N-1}


The first three terms n⋅KN⋅(1−KN)n \cdot \frac{K}{N} \cdot \left(1 - \frac{K}{N}\right) match the binomial variance formula with p=KNp = \frac{K}{N}.

The additional factor N−nN−1\frac{N-n}{N-1} is called the finite population correction and is always less than 1. It accounts for the reduction in variance caused by sampling without replacement. As the sample size nn approaches the population size NN, this factor approaches zero, reflecting that sampling the entire population leaves no variability.

Standard Deviation


σ=n⋅KN⋅N−KN⋅N−nN−1\sigma = \sqrt{n \cdot \frac{K}{N} \cdot \frac{N-K}{N} \cdot \frac{N-n}{N-1}}


Example


Consider drawing 5 cards from a standard deck of 52, counting aces. Here N=52N = 52, K=4K = 4, n=5n = 5:

Var(X)=5⋅452⋅4852⋅4751\mathrm{Var}(X) = 5 \cdot \frac{4}{52} \cdot \frac{48}{52} \cdot \frac{47}{51}


=5⋅113⋅1213⋅4751≈0.331= 5 \cdot \frac{1}{13} \cdot \frac{12}{13} \cdot \frac{47}{51} \approx 0.331


σ≈0.331≈0.575\sigma \approx \sqrt{0.331} \approx 0.575


The relatively small variance reflects the limited range of possible outcomes (0 to 4 aces) and the constraining effect of sampling without replacement from a finite deck.

Mode and Median

Mode


The mode is the value of kk (number of successes in the sample) with the highest probability—the peak of the PMF.

For the hypergeometric distribution, the mode depends on the parameters NN, KK, and nn:

The mode is ⌊(n+1)(K+1)N+2⌋\lfloor \frac{(n+1)(K+1)}{N+2} \rfloor

Intuition: The mode sits near the expected value nKN\frac{nK}{N}, representing the most likely number of successes when sampling without replacement. Unlike the binomial where trials are independent, the hypergeometric mode reflects how depletion of the population affects the probability distribution.

Example:
For N=50N = 50, K=20K = 20, n=10n = 10:

Mode = ⌊(10+1)(20+1)50+2⌋=⌊11×2152⌋=⌊4.44⌋=4\lfloor \frac{(10+1)(20+1)}{50+2} \rfloor = \lfloor \frac{11 \times 21}{52} \rfloor = \lfloor 4.44 \rfloor = 4

Getting exactly 4 successes is more likely than any other outcome.

Example:
For N=100N = 100, K=30K = 30, n=20n = 20:

Mode = ⌊21×31102⌋=⌊6.38⌋=6\lfloor \frac{21 \times 31}{102} \rfloor = \lfloor 6.38 \rfloor = 6

Median


The median is the value mm such that P(X≤m)≥0.5P(X \leq m) \geq 0.5 and P(X≥m)≥0.5P(X \geq m) \geq 0.5.

For the hypergeometric distribution, there is no simple closed-form expression for the median, but it can be found numerically by solving:

∑k=0m(Kk)(N−Kn−k)(Nn)≥0.5\sum_{k=0}^{m} \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}} \geq 0.5


Properties of the median:
• The median is always close to the mean nKN\frac{nK}{N}
• For large NN, the hypergeometric approaches the binomial, and median behavior converges
• The distribution can be symmetric or skewed depending on parameters

Example:
For N=50N = 50, K=25K = 25, n=10n = 10:
Mean = 10×2550=5\frac{10 \times 25}{50} = 5

The median is 5 (by symmetry, since K=N/2K = N/2)

Example:
For N=100N = 100, K=30K = 30, n=20n = 20:
Mean = 20×30100=6\frac{20 \times 30}{100} = 6

The median is approximately 6 (close to the mean)

Unlike continuous distributions where finding the median requires integration, for discrete distributions, the median is found by summing probabilities until reaching 0.5.

Applications and Examples


Practical Example


Suppose you have a deck of N=52N = 52 cards containing K=13K = 13 hearts. You draw n=5n = 5 cards without replacement. The probability of getting exactly k=2k = 2 hearts is:

P(X=2)=(132)(52−135−2)(525)=(132)(393)(525)=78⋅91392598960≈0.274P(X = 2) = \frac{\binom{13}{2} \binom{52-13}{5-2}}{\binom{52}{5}} = \frac{\binom{13}{2} \binom{39}{3}}{\binom{52}{5}} = \frac{78 \cdot 9139}{2598960} \approx 0.274


This means there's about a 27.4% chance of getting exactly 2 hearts when drawing 5 cards from a standard deck.

Note: When NN is very large relative to nn, the hypergeometric distribution approximates the binomial distribution with p=KNp = \frac{K}{N}.

Interactive Calculator


This interactive calculator computes probabilities for the hypergeometric distribution, which models sampling without replacement from a finite population. Enter your population size (NN), number of success items (KK), and sample size (nn) to calculate the probability of getting a specific number of successes in your sample. Ideal for card games, quality control sampling, lottery calculations, or any scenario where you're drawing items without putting them back.

1. Select probability type: 'All values' for full distribution, or choose a specific query like P(X=k) or P(X≤k)
2. Enter N (population size) - total items available
3. Enter K (success states) - how many items in population are 'successes'
4. Enter n (sample size) - how many items you're drawing
5. For specific queries, enter k (target successes in your sample)
6. Click Calculate to see probabilities and distribution

Hypergeometric Distribution Calculator

Calculate probabilities and distribution properties

Total number of items in population

Number of success items in population

Number of items drawn

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Hypergeometric at a Glance

The table below collects the full anatomy of the hypergeometric distribution into a single reference card — its three parameters, support bounds, the PMF and CDF, the mean and variance formulas (including the finite-population correction), mode and median behavior, the binomial approximation, and a canonical card-drawing example.
Aspect Formula / statement Note / example
Parameters N (population size), K (successes in population), n (sample size, n ≤ N) draws are made without replacement; trials are not independent
Support k ∈ {max(0, n − N + K), ..., min(n, K)} bounded by physical availability of successes and failures
PMF P(X = k) = C(K, k) · C(N − K, n − k) / C(N, n) favorable combinations over total combinations (Vandermonde's identity)
CDF F(k) = ∑i ≤ k C(K, i) · C(N − K, n − i) / C(N, n) no closed form; computed numerically
Expected value E[X] = n · K / N sample size × population proportion of successes
Variance Var(X) = n · (K/N) · ((N − K)/N) · ((N − n)/(N − 1)) binomial variance with p = K/N times the finite-population correction
Finite population correction (N − n) / (N − 1) — always ≤ 1; → 0 as n → N shrinks variance vs binomial; full census has zero variability
Mode and median mode = ⌊(n + 1)(K + 1) / (N + 2)⌋; median ≈ mean, found numerically both sit near nK/N; symmetric when K = N/2
Binomial approximation Hypergeometric(N, K, n) ≈ Binomial(n, K/N) when N ≫ n rule of thumb: n < 0.05·N — depletion becomes negligible
Canonical example draw 5 cards from 52, count hearts: N = 52, K = 13, n = 5 E[X] = 1.25; P(X = 2) ≈ 0.274