To understand the HyperGeometric distribution, consider a set of $$r$$ objects, of which $$m$$ are of the type I and $$n$$ are of the type II. A sample with size $$k$$ ($$k<r$$) with no replacement is randomly chosen. The number of observed type I elements observed in this sample is set to be our random variable $$X$$. For example, consider that in a set of 20 car parts, there are 4 that are defective (type I). If we take a sample of size 5 from those car parts, the probability of finding 2 that are defective will be given by the HyperGeometric distribution (needs double checking).

HyperGeometric(m, n, k)

## Arguments

m The number of type I elements available. The number of type II elements available. The size of the sample taken.

## Value

A HyperGeometric object.

## Details

We recommend reading this documentation on https://alexpghayes.github.io/distributions3, where the math will render with additional detail and much greater clarity.

In the following, let $$X$$ be a HyperGeometric random variable with success probability p = $$p = m/(m+n)$$.

Support: $$x \in { \{\max{(0, k-(n-m)}, \dots, \min{(k,m)}}\}$$

Mean: $$\frac{km}{n+m} = kp$$

Variance: $$\frac{km(n)(n+m-k)}{(n+m)^2 (n+m-1)} = kp(1-p)(1 - \frac{k-1}{m+n-1})$$

Probability mass function (p.m.f):

$$P(X = x) = \frac{{m \choose x}{n \choose k-x}}{{m+n \choose k}}$$

Cumulative distribution function (c.d.f):

$$P(X \le k) \approx \Phi\Big(\frac{x - kp}{\sqrt{kp(1-p)}}\Big)$$ Moment generating function (m.g.f):

Not useful.

Other discrete distributions: Bernoulli, Binomial, Categorical, Geometric, Multinomial, NegativeBinomial, Poisson

cdf(X, 4)#>  1quantile(X, 0.7)#>  4