Hypergeometric Distribution

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

Applications and Examples

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.

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.

Learn More About Distinguishing Discrete Distributions →

Example:

Drawing

5

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.

See More Examples of Hypergeometric Distribution →

Notation Used

X \sim \text{Hypergeometric}(N, K, n)

X \sim \text{Hyp}(N, K, n)

— distribution of the random variable.

\text{Hypergeometric}(N, K, n)

— used to denote the distribution itself (not the random variable).

\text{H}(N, K, n)

— occasionally used in compact form, especially in software or formulas.

P(X = k) = \frac{\binom{K}{k} \binom{N - K}{n - k}}{\binom{N}{n}}, \quad \text{for valid } k

— probability mass function (PMF), where:

N

— total population size

K

— number of success states in the population

n

— number of draws (sample size) without replacement

k

— number of observed successes in the sample

\max(0, n - (N - K)) \leq k \leq \min(n, K)

— valid range for

k

\binom{a}{b} = \frac{a!}{b!(a-b)!}

— binomial coefficient

Key properties:

E[X] = n \frac{K}{N}

— expected value (mean)

\text{Var}(X) = n \frac{K}{N} \frac{N-K}{N} \frac{N-n}{N-1}

— variance

Relationship to binomial distribution:

\text{Hypergeometric}(N, K, n) \approx \text{Binomial}(n, p)

where

p = \frac{K}{N}

, when

N

is large relative to

n

(sampling with replacement approximation)

See All Probability Symbols and Notations →

Parameters

𝑁

: total population size

𝐾

: number of successes in the population

𝑛

: number of draws (without replacement), where

𝑛≤𝑁

The hypergeometric distribution models the number of successes in

𝑛

draws from a finite population of size

𝑁

that contains exactly

𝐾

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) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}, \quad k = \max(0, n-N+K), \ldots, \min(n, K)

where

\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

n

items without replacement from a finite population of size

N

containing exactly

K

success items.

Support (Range of the Random Variable):
* The random variable

X

can take on values from

\max(0, n-N+K)

\min(n, K)

.
*

X = k

means exactly

k

successes are drawn in the sample of size

n

.
* The lower bound ensures we don't draw more failures than available:

n-k \leq N-K

* The upper bound ensures we don't draw more successes than available:

k \leq K

and

k \leq n

* The support is thus a finite set of non-negative integers.

Logic Behind the Formula:
*

\binom{K}{k}

: The number of ways to choose

k

successes from

K

available successes
*

\binom{N-K}{n-k}

: The number of ways to choose

n-k

failures from

N-K

available failures
*

\binom{N}{n}

: The total number of ways to choose

n

items from

N

items
* The total probability sums to 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

Population Size (N): 50

Success States (K): 20

Number of Draws (n): 10

Explanation

The hypergeometric distribution models the number of successes when sampling without replacement from a finite population. The probability mass function is $P(X = k) = \frac{\binom{K}{k} \binom{N-K}{n-k}}{\binom{N}{n}}$ , where $N$ is the population size, $K$ is the number of success states in the population, and $n$ is the number of draws. The expected value is $E[X] = n \cdot \frac{K}{N}$ and the variance is $\text{Var}(X) = n \cdot \frac{K}{N} \cdot \left(1-\frac{K}{N}\right) \cdot \frac{N-n}{N-1}$ . 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:

F_X(k) = P(X \leq k) = \sum_{i=0}^{k} \frac{\binom{K}{i}\binom{N-K}{n-i}}{\binom{N}{n}}

Where:

N

= total population size

K

= number of success states in the population

n

= number of draws (sample size)

k

= number of observed successes in the sample (where

\max(0, n-N+K) \leq k \leq \min(n, K)

)

Intuition Behind the Formula

Definition: The CDF gives the probability of observing

k

or fewer successes when drawing

n

items without replacement from a population of size

N

containing

K

success states.

Summation of Probabilities:

We sum the individual probabilities from the minimum possible value up to

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:

$\max(0, n-N+K)$ (can't draw more failures than exist)
$\min(n, K)$ (can't draw more successes than exist or more items than drawn)

Complementary Probability:

For "more than

k

successes":

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

Hypergeometric Distribution CDF

CDF for sampling without replacement

Population Size (N): 50

Success States (K): 20

Number of Draws (n): 10

CDF Explanation

The hypergeometric CDF is

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)) \leq k \leq \min(n, K)

. This represents the probability of drawing

k

or fewer success items when sampling

n

items without replacement from a population of size

N

containing

K

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] = \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 \cdot \frac{K}{N}

Where:

N

= total population size

K

= number of success states in the population

n

= 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

\frac{K}{N}

. When drawing

n

items, each draw has the same marginal probability

\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

n

draws is:

E[X] = n \cdot \frac{K}{N}

The result

E[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

n

items from a population where the success rate is

\frac{K}{N}

, you expect

n \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 = 52

K = 4

, and

n = 5

E[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:

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

Or using the shortcut formula:

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

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

Formula

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

Where:

N

= total population size

K

= number of success states in the population

n

= number of draws (sample size)

Derivation and Intuition

The derivation involves computing

\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

\mu = n \cdot \frac{K}{N}

.

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

\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 \cdot \frac{K}{N} \cdot \left(1 - \frac{K}{N}\right)

match the binomial variance formula with

p = \frac{K}{N}

.

The additional factor

\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

n

approaches the population size

N

, this factor approaches zero, reflecting that sampling the entire population leaves no variability.

Standard Deviation

\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 = 52

K = 4

n = 5

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

= 5 \cdot \frac{1}{13} \cdot \frac{12}{13} \cdot \frac{47}{51} \approx 0.331

\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.

Applications and Examples

Practical Example

Suppose you have a deck of

N = 52

cards containing

K = 13

hearts. You draw

n = 5

cards without replacement. The probability of getting exactly

k = 2

hearts is:

P(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

N

is very large relative to

n

, the hypergeometric distribution approximates the binomial distribution with

p = \frac{K}{N}