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Hypergeometric Distribution Formula

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The hypergeometric distribution is defined as the concept of approximation of a random variable in a hypergeometric probability distribution. This value is further used to evaluate the probability distribution function of the data. The hypergeometric distribution resembles the binomial distribution in terms of a probability distribution. Combinations and binomial distribution are employed in hypergeometric distribution to do the calculations. It is used to determine statistical measures such as mean, standard deviation, and variance.

Hypergeometric Distribution Formula

The formula for Hypergeometric Distribution is given by,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}

where,

  • P(x | N, m, n) is the hypergeometric probability for exactly x successes when population consists of N items out of which m are successes,
  • x is the number of items in the sample known as successes,
  • N is the number of items in the population,
  • n is the number of items in the sample,
  • m is the number of successes in the sample.

Sample Problems

Problem 1: Find the probability density function of the hypergeometric function if the values of N, n and m are 40, 20 and 10 respectively.

Solution:

We have,

N = 40

n = 20

m = 10

Using the formula we get,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}\\ P (x | N, m, n) = \frac{\left(\binom{10}{x}\binom{40-10}{20-x}\right)}{\binom{40}{20}}\\ P (x | N, m, n) = \frac{\left(\binom{10}{x}\binom{30}{20-x}\right)}{\binom{40}{20}}

Problem 2: Find the probability density function of the hypergeometric function if the values of N, n, and m are 70, 30, and 15 respectively.

Solution:

We have,

N = 70

n = 30

m = 15

Using the formula we get,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}\\ P (x | N, m, n) = \frac{\left(\binom{15}{x}\binom{70-15}{30-x}\right)}{\binom{70}{30}}\\ P (x | N, m, n) = \frac{\left(\binom{15}{x}\binom{55}{30-x}\right)}{\binom{70}{30}}

Problem 3: Find the probability density function of the hypergeometric function if the values of N, n and m are 30, 5 and 3 respectively.

Solution:

We have,

N = 30

n = 5

m = 3

Using the formula we get,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}\\ P (x | N, m, n) = \frac{\left(\binom{3}{x}\binom{30-3}{5-x}\right)}{\binom{30}{3}}\\ P (x | N, m, n) = \frac{\left(\binom{3}{x}\binom{27}{5-x}\right)}{\binom{30}{3}}

Problem 4: Find the probability density function of the hypergeometric function if the values of N, n and m are 60, 25 and 20 respectively.

Solution:

We have,

N = 60

n = 25

m = 20

Using the formula we get,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}\\ P (x | N, m, n) = \frac{\left(\binom{20}{x}\binom{60-20}{25-x}\right)}{\binom{60}{25}}\\ P (x | N, m, n) = \frac{\left(\binom{20}{x}\binom{40}{25-x}\right)}{\binom{60}{25}}

Problem 5: Find the probability density function of the hypergeometric function if the values of N, n and m are 100, 60 and 50 respectively.

Solution:

We have,

N = 100

n = 60

m = 50

Using the formula we get,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}\\ P (x | N, m, n) = \frac{\left(\binom{50}{x}\binom{100-50}{60-x}\right)}{\binom{100}{60}}\\ P (x | N, m, n) = \frac{\left(\binom{50}{x}\binom{50}{60-x}\right)}{\binom{100}{60}}

Problem 6: Find the probability density function of the hypergeometric function if the values of N, n and m are 200, 40 and 30 respectively.

Solution:

We have,

N = 200

n = 40

m = 30

Using the formula we get,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}\\ P (x | N, m, n) = \frac{\left(\binom{30}{x}\binom{200-30}{40-x}\right)}{\binom{200}{40}}\\ P (x | N, m, n) = \frac{\left(\binom{30}{x}\binom{170}{40-x}\right)}{\binom{200}{40}}

Problem 7: Find the probability density function of the hypergeometric function if the values of N, n and m are 70, 20 and 15 respectively.

Solution:

We have,

N = 70

n = 20

m = 15

Using the formula we get,

P (x | N, m, n) = \frac{\left(\binom{m}{x}\binom{N-m}{n-x}\right)}{\binom{N}{n}}\\ P (x | N, m, n) = \frac{\left(\binom{15}{x}\binom{70-15}{20-x}\right)}{\binom{70}{20}}\\ P (x | N, m, n) = \frac{\left(\binom{15}{x}\binom{55}{20-x}\right)}{\binom{70}{20}}



Last Updated : 06 Jan, 2024
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