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## Hypergeometric Distribution

The probability distribution of a hypergeometric random variable is called a **hypergeometric distribution.** This lesson describes how hypergeometric random variables, hypergeometric experiments, hypergeometric probability, and the hypergeometric distribution are all related.

## Notation

The following notation is helpful, when we talk about hypergeometric distributions and hypergeometric probability.

*N*: The number of items in the population.*k*: The number of items in the population that are classified as successes.*n*: The number of items in the sample.*x*: The number of items in the sample that are classified as successes._{k}C_{x}: The number of combinations of*k*things, taken*x*at a time.- h(
*x*;*N*,*n*,*k*):**hypergeometric probability**– the probability that an*n*-trial hypergeometric experiment results in exactly*x*successes, when the population consists of*N*items,*k*of which are classified as successes.

## Hypergeometric Experiments

A **hypergeometric experiment** is a statistical experiment that has the following properties:

- A sample of size
*n*is randomly selected without replacement from a population of*N*items. - In the population,
*k*items can be classified as successes, and*N – k*items can be classified as failures.

Consider the following statistical experiment. You have an urn of 10 marbles – 5 red and 5 green. You randomly select 2 marbles without replacement and count the number of red marbles you have selected. This would be a hypergeometric experiment.

Note that it would not be a binomial experiment. A binomial experiment requires that the probability of success be constant on every trial. With the above experiment, the probability of a success changes on every trial. In the beginning, the probability of selecting a red marble is 5/10. If you select a red marble on the first trial, the probability of selecting a red marble on the second trial is 4/9. And if you select a green marble on the first trial, the probability of selecting a red marble on the second trial is 5/9.

Note further that if you selected the marbles with replacement, the probability of success would not change. It would be 5/10 on every trial. Then, this would be a binomial experiment.

## Hypergeometric Distribution

A **hypergeometric random variable** is the number of successes that result from a hypergeometric experiment. The probability distribution of a hypergeometric random variable is called a **hypergeometric distribution**.

Given *x*, *N*, *n*, and *k*, we can compute the hypergeometric probability based on the following formula:

**Hypergeometric Formula.**. Suppose a population consists of *N* items, *k* of which are successes. And a random sample drawn from that population consists of *n* items, *x* of which are successes. Then the hypergeometric probability is:

h(*x*; *N*, *n*, *k*) = [ _{k}C_{x} ] [ _{N-k}C_{n-x} ] / [ _{N}C_{n} ]

The hypergeometric distribution has the following properties:

- The mean of the distribution is equal to
*n***k*/*N*. - The variance is
*n***k** (*N*–*k*) * (*N*–*n*) / [*N*^{2}* (*N*– 1 ) ] .

**Example 1**

Suppose we randomly select 5 cards without replacement from an ordinary deck of playing cards. What is the probability of getting exactly 2 red cards (i.e., hearts or diamonds)?

*Solution:* This is a hypergeometric experiment in which we know the following:

- N = 52; since there are 52 cards in a deck.
- k = 26; since there are 26 red cards in a deck.
- n = 5; since we randomly select 5 cards from the deck.
- x = 2; since 2 of the cards we select are red.

We plug these values into the hypergeometric formula as follows:

h(*x*; *N*, *n*, *k*) = [ _{k}C_{x} ] [ _{N-k}C_{n-x} ] / [ _{N}C_{n} ]

h(*2*; *52*, *5*, *26*) = [ _{26}C_{2} ] [ _{26}C_{3} ] / [ _{52}C_{5} ]

h(*2*; *52*, *5*, *26*) = [ 325 ] [ 2600 ] / [ 2,598,960 ]

h(*2*; *52*, *5*, *26*) = 0.32513

Thus, the probability of randomly selecting 2 red cards is 0.32513.

## Cumulative Hypergeometric Probability

A **cumulative hypergeometric probability** refers to the probability that the hypergeometric random variable is greater than or equal to some specified lower limit and less than or equal to some specified upper limit.

For example, suppose we randomly select five cards from an ordinary deck of playing cards. We might be interested in the cumulative hypergeometric probability of obtaining 2 or fewer hearts. This would be the probability of obtaining 0 hearts plus the probability of obtaining 1 heart plus the probability of obtaining 2 hearts, as shown in the example below.

**Example 1**

Suppose we select 5 cards from an ordinary deck of playing cards. What is the probability of obtaining 2 or fewer hearts?

*Solution:* This is a hypergeometric experiment in which we know the following:

- N = 52; since there are 52 cards in a deck.
- k = 13; since there are 13 hearts in a deck.
- n = 5; since we randomly select 5 cards from the deck.
- x = 0 to 2; since our selection includes 0, 1, or 2 hearts.

We plug these values into the hypergeometric formula as follows:

h(*x* < x; *N*, *n*, *k*) = h(*x* < 2; *52*, *5*, *13*)

h(*x* < 2; *52*, *5*, *13*) = h(*x* = 0; *52*, *5*, *13*) + h(*x* = 1; *52*, *5*, *13*) + h(*x* = 2; *52*, *5*, *13*)

h(*x* < 2; *52*, *5*, *13*) = [ (_{13}C_{0}) (_{39}C_{5}) / (_{52}C_{5}) ] + [ (_{13}C_{1}) (_{39}C_{4}) / (_{52}C_{5}) ] + [ (_{13}C_{2}) (_{39}C_{3}) / (_{52}C_{5}) ]

h(*x* < 2; *52*, *5*, *13*) = [ (1)(575,757)/(2,598,960) ] + [ (13)(82,251)/(2,598,960) ] + [ (78)(9139)/(2,598,960) ]

h(*x* < 2; *52*, *5*, *13*) = [ 0.2215 ] + [ 0.4114 ] + [ 0.2743 ]

h(*x* < 2; *52*, *5*, *13*) = 0.9072

Thus, the probability of randomly selecting at most 2 hearts is 0.9072.

## Hypergeometric Distribution

and the kurtosis excess is given by a complicated expression.

The generating function is

## Hypergeometric Distribution Definition

In the statistics and the probability theory, hypergeometric distribution is basically a distinct probability distribution which defines probability of k successes (i.e. some random draws for the object drawn that has some specified feature) in n no of draws, without any replacement, from a given population size N which includes accurately K objects having that feature, where the draw may succeed or may fail.

The formula for the probability of a hypergeometric distribution is derived using a number of items in the population, number of items in the sample, number of successes in the population, number of successes in the sample, and few combinations. Mathematically, the probability is represented as,

**P = _{K }C _{k }* _{(N – K) }C _{(n – k) }/_{ N }C _{n}**

where,

- N = No. of items in the population
- n = No. of items in the sample
- K = No. of successes in the population
- k = No. of successes in the sample

The mean and standard deviation of a hypergeometric distribution is expressed as,

**Mean = n * K / NStandard Deviation = [n * K * (N – K) * (N – n) / {N ^{2} * (N – 1)}]^{1/2}**

### Explanation

Follow the below steps:

**Firstly, determine the total number of items in the population, which is denoted by N. For example, the number of playing cards in a deck is 52.****Next, determine the number of items in the sample, denoted by n—for example, the number of cards drawn from the deck.****Next, determine the instances which will be considered to be successes in the population, and it is denoted by K. For example, the number of hearts in the overall deck, which is 13.****Next, determine the instances which will be considered to be successes in the sample drawn, and it is denoted by k. E.g., the number of hearts in the cards drawn from the deck.****Finally, the formula for the probability of a hypergeometric distribution is derived using a number of items in the population (step 1), number of items in the sample (step 2), number of successes in the population (step 3) and number of successes in the sample (step 4) as shown below.**

**P = _{K }C _{k }* _{(N – K) }C _{(n – k) }/_{ N }C _{n}**

## Examples of Hypergeometric Distribution (with Excel Template)

#### Example #1

**Let us take the example of an ordinary deck of playing cards form where 6 cards are drawn randomly without replacement. Determine the probability of drawing exactly 4 red suites cards, i.e., diamonds or hearts.**

**Given, N = 52 (since there are 52 cards in an ordinary playing deck)****n = 6 (Number of cards drawn randomly from the deck)****K = 26 (since there are 13 red cards each in diamonds and hearts suite)****k = 4 (Number of red cards to be considered successful in the sample drawn)**

**Solution:**

Therefore, the probability of drawing exactly 4 red suites cards in the drawn 6 cards can be calculated using the above formula as,

Probability = _{K }C _{k }* _{(N – K) }C _{(n – k) }/_{ N }C _{n}

= _{26 }C _{4 }* _{(52 – 26) }C _{(6 – 4) }/_{ 52 }C _{6}

= _{26 }C _{4 }* _{26 }C _{2 }/_{ 52 }C _{6}

= 14950 * 325 / 20358520

The probability will be –

**Probability = 0.2387 ~ 23.87%**

Therefore, there is a 23.87% probability of drawing exactly 4 red cards while drawing 6 random cards from an ordinary deck.

#### Example #2

**Let us take another example of a wallet that contains 5 $100 bills and 7 $1 bills. If 4 bills are chosen randomly, then determine the probability of choosing exactly 3 $100 bills.**

**Given, N = 12 (Number of $100 bills + Number of $1 bills)****n = 4 (Number of bills chosen randomly)****K = 5 (since there are 5 $100 bills)****k = 3 (Number of $100 bills to be considered a success in the sample chosen)**

**Solution:**

Therefore, the probability of choosing exactly 3 $100 bills in the randomly chosen 4 bills can be calculated using the above formula as,

Probability = _{K }C _{k }* _{(N – K) }C _{(n – k) }/_{ N }C _{n}

= _{5 }C _{3 }* _{(12 – 5) }C _{(4 – 3) }/_{ 12 }C _{4}

= _{5 }C _{3 }* _{7 }C _{1 }/_{ 12 }C _{4}

= 10 * 7 / 495

Probability will be –

**Probability = 0.1414 ~ 14.14%**

Therefore, there is a 14.14% probability of choosing exactly 3 $100 bills while drawing 4 random bills.

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