# Hypergeometric Distribution

## Hypergeometric Distribution

A hypergeometric experiment is an experiment which satisfies each of the following conditions:

• The population or set to be sampled consists of $N$ individuals, objects, or elements (a finite population).
• Each object can be characterized as a "defective" or "non-defective", and there are $M$ defectives in the population.
• A sample of $n$ individuals is drawn in such a way that each subset of size $n$ is equally likely to be chosen.

## Hypergeometric Distribution

Suppose we have an hypergeometric experiment. That is, suppose there are $N$ units in the population and $M$ out of $N$ are defective, so $N-M$ units are non-defective.

Let $X$ denote the number of defective in a completely random sample of size $n$ drawn from a population consisting of total $N$ units.

The total number of ways of finding $n$ units out of $N$ is $\binom{N}{n}$.

Out of $M$ defective units $x$ defective units can be selected in $\binom{M}{x}$ ways and out of $N-M$ non-defective units remaining $(n-x)$ units can be selected in $\binom{N-M}{n-x}$ ways.

Hence, probability of selecting $x$ defective units in a random sample of $n$ units out of $N$ is

 $$\begin{equation*} P(X=x) =\frac{\text{Favourable Cases}}{\text{Total Cases}} \end{equation*}$$

 $$\begin{equation*} \therefore P(X=x)=\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}},\;\; x=0,1,2,\cdots, n. \end{equation*}$$

The above distribution is called hypergeometric distribution.

Notation: $X\sim H(n,M,N)$.

## Graph of Hypergeometric Distribution H(5,5,20)

Following graph shows the probability mass function of hypergeometric distribution.

## Key Features of Hypergeometric Distribution

• Suppose there are $N$ units in the population. These $N$ units are classified as $M$ successes and remaining $N-M$ failures.
• Out of $N$ units, $n$ units are selected at random without replacement.
• $X$ is the number of successes in the sample.

## Mean of Hypergeometric Distribution

The expected value of hypergeometric randome variable is $E(X) =\dfrac{Mn}{N}$.

#### Proof

The expected value of hypergeometric randome variable is

 $$\begin{eqnarray*} E(x) &=& \sum_{x=0}^n x\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}\\ &=& 0+ \sum_{x=1}^n x\frac{\frac{M!}{x!(M-x)!}\binom{N-M}{n-x}}{\frac{N!}{n!(N-n)!}}\\ &=& \sum_{x=1}^n \frac{\frac{M(M-1)!}{(x-1)!(M-x)!}\binom{N-M}{n-x}}{\frac{N(N-1)!}{n(n-1)!(N-n)!}}\\ &=& \frac{Mn}{N}\sum_{x=1}^n\frac{\binom{M-1}{x-1}\binom{N-M}{n-x}}{\binom{N-1}{n-1}} \end{eqnarray*}$$

Let $x-1=y$. So for $x=1$, $y=0$ and for $x=n$, $y=n-1$. Therefore

 $$\begin{eqnarray*} \mu_1^\prime &=& \frac{Mn}{N}\sum_{y=0}^{n-1}\frac{\binom{M-1}{y}\binom{N-M}{n-y-1}}{\binom{N-1}{n-1}} \\ &=& \frac{Mn}{N}\sum_{y=0}^{n^\prime}\frac{\binom{M-1}{y}\binom{N-M}{n^\prime-y}}{\binom{N-1}{n^\prime-1}} \\ &=&\frac{Mn}{N}\times 1. \end{eqnarray*}$$

Hence, mean = $E(X) =\dfrac{Mn}{N}$.

## Variance of Hypergeometric Distribution

The variance of an hypergeometric random variable is $V(X) = \dfrac{Mn(N-M)(N-n)}{N^2(N-1)}$.

#### Proof

The variance of random variable $X$ is given by

 $$\begin{equation*} V(X) = E(X^2) - [E(X)]^2. \end{equation*}$$

Let us find the expected value of $X(X-1)$.

 $$\begin{eqnarray*} E[X(X-1)]&=& \sum_{x=0}^n x(x-1)\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}\\ &=& 0+0+ \sum_{x=2}^n x\frac{\frac{M!}{x!(M-x)!}\binom{N-M}{n-x}}{\frac{N!}{n!(N-n)!}}\\ &=& \sum_{x=2}^n \frac{\frac{M(M-1)(M-2)!}{(x-2)!(M-x)!}\binom{N-M}{n-x}}{\frac{N(N-1)(N-2)!}{n(n-1)(n-2)!(N-n)!}}\\ &=& \frac{M(M-1)n(n-1)}{N(N-1)}\sum_{x=2}^n\frac{\binom{M-2}{x-2}\binom{N-M}{n-x}}{\binom{N-2}{n-2}} \end{eqnarray*}$$

Let $x-2=y$. So for $x=2$, $y=0$ and for $x=n$, $y=n-2$. Therefore

 $$\begin{eqnarray*} E[(X(X-1)]&=& \frac{Mn}{N}\sum_{y=0}^{n-2}\frac{\binom{M-2}{y}\binom{N-M}{n-y-2}}{\binom{N-2}{n-2}} \\ &=& \frac{Mn}{N}\sum_{y=0}^{n^\prime}\frac{\binom{M-2}{y}\binom{N-M}{n^\prime-y}}{\binom{N-2}{n^\prime}} \\ &=& \frac{M(M-1)n(n-1)}{N(N-1)}\times 1\\ & = &\frac{M(M-1)n(n-1)}{N(N-1)}. \end{eqnarray*}$$

The second raw moment is given by

 $$\begin{eqnarray*} \mu_2^\prime &=& E[X(X-1)]+E(X) \\ &=& \frac{M(M-1)n(n-1)}{N(N-1)}+ \frac{Mn}{N}. \end{eqnarray*}$$

Hence, the variance of hypergrometric distribution is

 $$\begin{eqnarray*} \text{Variance = }\mu_2 &=& \mu_2^\prime -(\mu_1^\prime)^2 \\ &=& \frac{M(M-1)n(n-1)}{N(N-1)}+ \frac{Mn}{N}- \frac{M^2n^2}{N^2} \\ &=& \frac{Mn(N-M)(N-n)}{N^2(N-1)}. \end{eqnarray*}$$

## Binomial as a limiting case of Hypergeometric distribution

In Hypergeometric distribution, if $N\to \infty$ and $\frac{M}{N}=p$, then the hypergeometric distribution tends to binomial distribution.

#### Proof

The probability mass function of hypergeometric distribution is

 $$\begin{equation*} \therefore P(X=x)=\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}},\;\; x=0,1,2,\cdots, n. \end{equation*}$$

Taking limit as $N\to \infty$, we have

 $$\begin{eqnarray*} P(X=x) &=& \lim_{N\to\infty} \frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}\\ &=& \lim_{N\to\infty} \frac{\bigg[\frac{M(M-1)\cdots (M-x+1)}{x!}\bigg]\bigg[\frac{(N-M)(N-M-1)\cdots (N-M-n+x+1)}{(n-x)!}\bigg]}{\frac{N(N-1)\cdots (N-n+1)}{n!}} \end{eqnarray*}$$

Dividing numerator and denominator by $N$, we get

 $$\begin{eqnarray*} & & P(X=x)\\ &=& \lim_{N\to\infty} \frac{n!}{x!(n-x)!}\frac{\frac{M}{N}(\frac{M}{N}-\frac{1}{N})\cdots (\frac{M}{N}-\frac{x-1}{N})(1-\frac{M}{N})(1-\frac{M}{N}-\frac{1}{N})\cdots (1-\frac{M}{N}-\frac{n-x-1}{N})}{1(1-\frac{1}{N})\cdots (1-\frac{n-1}{N})}\\ & = &\binom{n}{x}\frac{p(p-0)\cdots (p-0)(1-p)(1-p-0)\cdots (1-p-0)}{1(1-0)\cdots (1-0)}\;\;\; (\because \frac{M}{N}=p)\\ & = &\binom{n}{x}p^x (1-p)^{n-x}, x=0,1,2,\cdots, n; \; 0 < p < 1. \end{eqnarray*}$$

which is the probability mass function of binomial distribution.

Hope this tutorial helps you understand Hypergeometric distribution and various results related to Hypergeometric distributions. VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.