For discrete random variables, besides binomial and multinomial distributions, there are other ones. In this page, we examine two of them: hypergeometric and Poisson distributions.
Hypergeometric distribution
Consider a set of `N` elements with `m` elements belong to group S. Take a sample of `n` elements without replacement. Then probability that `x` elements belong to group S is determined by:
| `p(x)=P(X=x)=(((m),(x))((N-m),(n-x)))/(((N),(n)))=((m!)/(x!(m-x)!)\ ((N-m)!)/((n-x)!(N-m-n+x)!))/((N!)/(n!(N-n)!))` | (4) |
`x` must be in the interval `max(0,n+m–N)` and `min(m,n)`.
We can recognize that the basic difference of binomial and hypergeometric distributions is the method of sampling. In hypergeometric distribution, sample is taken without replacement, where as in binomial distribution, sample is taken with replacement.
Example 7 : In a box, there are 8 products of company A and 12 products of other companies. Take 8 products in succession without replacement from this box. What is the probability that there are 3 products of company A among 8 products?
We use hypergeometric distribution with `N=20` ; `m=8` ; `n=8` ; `x=3` . Then:
`((m),(x))=((8),(3))=(8!)/(3! xx5!)=56`
`((N-m),(n-x))=((12),(5))=(12!)/(5! xx7!)=792`
`((N),(n))=((20),(8))=(20!)/(8! xx12!)=125.970`
Therefore :
`p(3)=P(X=3)=(56xx972)/125.970=0,3521`
Note : When the number of elements of sample `n` is small in comparison with that of set `N` (`n< N//10`), hypergeometric distribution is approximative to binomial one with `p=m//N`.
Poisson distribution
Considering a group which the probability of property S is `p`. Realizing `n` independent trials from this group. Upon to large number rule, there are `lambda` times whose property are S (`lambda=np`). The probability that there are `x` times whose property are S is:
| `p(x)=P(X=x)=(e^(-lambda)lambda^x)/(x!)` | (11) |
This distribution is denoted as Poisson distribution and very useful when `n` is large and `p` is small.
Example 8 : 2% of products of company C are defective. Take 200 products (with replacement); what is the probability that there are 5 defective products?
By large number rule, in 200 products, there are `lambda=200xx0,02=4` defective products.
Apply Poisson distribution, we get:
`P(X=5)=(e^(-4)xx4^5)/(5!)=0,156`
Note : Poisson problems can be met in the following form:
Consider a discrete random variable `X` following Poisson distribution with the mean `lambda`. The probability that this variable get the value `x` is:
| `p(x)=P(X=x)=(e^(-lambda)lambda^x)/(x!)` | (11b) |
Example 9 : The number of defective products in one day of factory F follows Poisson distribution with mean of 3. The probability that there are 4 defective products in one day is:
`P(X=4)=(e^(-3)xx3^4)/(4!)=0,168`
Note : We can use Poisson distribution instead of binomial one when `np<=5` or `n(1–p)<=5`.
