Hypergeometric distribution finds its use in both probability and statistical theory. A hypergeometric random variable denotes the total number of successes in a hypergeometric experiment. It denotes to the experiments containing number of draws without replacement. For example, if $5$ cards are chosen from a deck of $52$ cards without replacement then it will be called a hypergeometric experiment. Cumulative hypergeometric probability comes when the total number of successes will be greater than a lower limit or lower than an upper limit. For example, if $5$ cards are chosed from a deck of $52$ cards without replacement then the probability of getting at least two kings will be a cumulative probability.

## Definition

Hypergeometric distribution is the probability distribution of a hypergeometric random variable. Hypergeometric distribution can be defined as discrete probability distribution describing the probability of getting $k$ successes in $n$ draws without replacement where the sample population is $N$. Each draw is either success or failure and the population consists of exactly $K$ successes. In statistics, the hypergeometric test uses the hypergeometric distribution to see the effect of $k$ successes statistically.

Properties of a hypergeometric experiment are:

**1)**A sample size of $n$ is collected without replacement from a population of $N$.

**2)**If $k$ items in the population $N$ is defined as success, then $(N - k)$ items will be defined as failures.

## Formula

The probability mass function of a random variable

**X**having hypergeometric distribution is given as,$\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$

$N$ = Population

$n$ = Sample or number of draws

$K$ = Number of success in population $N$

$k$ = Number of success came in the draws

$\binom{a}{b}$ = Binomial Coefficient

## Mean & Variance

Variance of a hypergeometric function is given by the formula: $\frac{n\times k\times (N-k)\times (N-n)}{N^2\times (N-1)}$.

Let us take an example and understand. There are $52$ cards in a deck. Find the probability of getting $1$ red card out of two cards chosen randomly without replacement.

We have, total population = $52$

sample population, $n$ = $2$

number of successes in total population, $K$ = $26$

number of successes in draws, $k$ = $1$

Now, the probability of getting one success will be,

$P(X = 1)$ = $\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}$ = $\frac{\binom{26}{1}\binom{26}{1}}{\binom{52}{2}}$

= $\frac{26}{26\times 26\times 51}$ = $0.000754$