HYPERGEOMETRIC DISTRIBUTION: FORMULAS, EQUATIONS, MODEL - DUDAS - 2025

The hypergeometric distribution is a discrete statistical function, suitable for calculating the probability in randomized experiments with two possible outcomes. The condition that is required to apply it is that they are small populations, in which the withdrawals are not replaced and the probabilities are not constant.

Therefore, when an element of the population is chosen to know the result (true or false) of a certain characteristic, that same element cannot be chosen again.

Figure 1. In a bolt population like this, there are surely defective specimens. Source: Pixabay.

Certainly, the next element chosen is thus more likely to obtain a true result, if the previous element had a negative result. This means that the probability varies as elements are extracted from the sample.

The main applications of the hypergeometric distribution are: quality control in processes with little population and the calculation of probabilities in games of chance.

As for the mathematical function that defines the hypergeometric distribution, it consists of three parameters, which are:

- Number of population elements (N)

- Sample size (m)

- Number of events in the entire population with a favorable (or unfavorable) result of the studied characteristic (n).

Formulas and equations

The formula for the hypergeometric distribution gives the probability P that x favorable cases of a certain characteristic occur. The way to write it mathematically, based on the combinatorial numbers is:

In the previous expression N, n and m are parameters and x is the variable itself.

- Total population is N.

-Number of positive results of a certain binary characteristic with respect to the total population is n.

-Quantity of elements in the sample is m.

In this case, X is a random variable that takes the value x and P (x) indicates the probability of occurrence of x favorable cases of the characteristic studied.

Important statistical variables

Other statistical variables for the hypergeometric distribution are:

- Mean μ = m * n / N

- Variance σ ^ 2 = m * (n / N) * (1-n / N) * (Nm) / (N-1)

- Standard deviation σ which is the square root of the variance.

Model and properties

To arrive at the model of the hypergeometric distribution, we start from the probability of obtaining x favorable cases in a sample of size m. This sample contains elements that comply with the property under study and elements that do not.

Recall that n represents the number of favorable cases in the total population of N elements. Then the probability would be calculated like this:

Expressing the above in the form of combinatorial numbers, the following probability distribution model is reached:

Main properties of the hypergeometric distribution

They are as follows:

- The sample must always be small, even if the population is large.

- The elements of the sample are extracted one by one, without incorporating them back into the population.

- The property to be studied is binary, that is, it can only take two values: 1 or 0, or true or false.

In each element extraction step, the probability changes depending on the previous results.

Approximation using the binomial distribution

Another property of the hypergeometric distribution is that it can be approximated by the binomial distribution, denoted Bi, as long as the population N is large and at least 10 times larger than the sample m. In this case it would look like this:

The probability that x = 3 screws in the sample are defective is: P (500, 5, 60, 3) = 0.0129.

For its part, the probability that x = 4 screws out of the sixty of the sample are defective is: P (500, 5, 60; 4) = 0.0008.

Finally, the probability that x = 5 screws in that sample are defective is: P (500, 5, 60; 5) = 0.

But if you want to know the probability that in that sample there are more than 3 defective screws, then you have to obtain the cumulative probability, adding:

This example is illustrated in figure 2, obtained by using GeoGebra, a free software widely used in schools, institutes and universities.

Figure 2. Example of hypergeometric distribution. Prepared by F. Zapata with GeoGebra.

Example 2

A Spanish deck deck has 40 cards, of which 10 have gold and the remaining 30 do not. Suppose that 7 cards are drawn at random from that deck, which are not reincorporated into the deck.

If X is the number of golds present in the 7 cards drawn, then the probability of having x golds in a 7-card draw is given by the hypergeometric distribution P (40,10,7; x).

Let's see this like this: to calculate the probability of having 4 golds in a 7-card draw we use the formula of the hypergeometric distribution with the following values:

And the result is: 4.57% probability.

But if you want to know the probability of getting more than 4 cards, then you have to add:

Solved exercises

The following set of exercises is intended to illustrate and assimilate the concepts that have been presented in this article. It is important that the reader tries to solve them on his own, before looking at the solution.

Exercise 1

A condom factory has found that out of every 1000 condoms produced by a certain machine, 5 are defective. For quality control, 100 condoms are taken at random and the lot is rejected if there is at least one or more defective. Answer:

a) What is the possibility that a lot of 100 will be discarded?

b) Is this quality control criterion efficient?

Solution

In this case, very large combinatorial numbers will appear. The calculation is difficult, unless you have a suitable software package.

But since it is a large population and the sample is ten times smaller than the total population, it is possible to use the approximation of the hypergeometric distribution by the binomial distribution:

In the above expression C (100, x) is a combinatorial number. Then the probability of having more than one defective will be calculated like this:

It is an excellent approximation, if compared to the value obtained by applying the hypergeometric distribution: 0.4102

It can be said that, with a 40% probability, a batch of 100 prophylactics should be discarded, which is not very efficient.

But, being a little less demanding in the quality control process and discarding the lot of 100 only if there are two or more defectives, then the probability of discarding the lot would fall to just 8%.

Exercise 2

A plastic block machine works in such a way that out of every 10 pieces, one comes out deformed. In a sample of 5 pieces, how likely is it that only one piece is defective?

Solution

Population: N = 10

Number n of defectives for every N: n = 1

Sample size: m = 5

Therefore there is a 50% probability that in a sample of 5, a block will be deformed.

Exercise 3

In a meeting of young high school graduates there are 7 ladies and 6 gentlemen. Among the girls, 4 study humanities and 3 science. In the boy group, 1 studies humanities and 5 science. Calculate the following:

a) Choosing three girls at random: how likely is it that they all study humanities?

b) If three attendees to the friends' meeting are chosen at random: What is the possibility that three of them, regardless of gender, study science all three, or humanities also all three?

c) Now select two friends at random and call x the random variable "number of those who study humanities". Between the two chosen, determine the mean or expected value of x and the variance σ ^ 2.

Solution to

The values ​​to use now are:

-Population: N = 14

-Quantity that studies letters is: n = 6 and the

-Size of the sample: m = 3.

-Number of friends studying humanities: x

According to this, x = 3 means that all three study humanities, but x = 0 means that none study humanities. The probability that all three study the same is given by the sum:

P (14, 6, 3, x = 0) + P (14, 6, 3, x = 3) = 0.0560 + 0.1539 = 0.2099

Then we have a 21% probability that three meeting attendees, chosen at random, will study the same thing.

Solution c

Here we have the following values:

N = 14 total population of friends, n = 6 total number in the population studying humanities, the sample size is m = 2.

Hope is:

E (x) = m * (n / N) = 2 * (6/14) = 0.8572

And the variance:

σ (x) ^ 2 = m * (n / N) * (1-n / N) * (Nm) / (N-1) = 2 * (6/14) * (1-6 / 14) * (14-2) / (14-1) =

= 2 * (6/14) * (1-6 / 14) * (14-2) / (14-1) = 2 * (3/7) * (1-3 / 7) * (12) / (13) = 0.4521

References

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