Consider a statistical experiment that there are only two outcomes (or reduced to two outcomes) S and F (success and failure). The probabilities of S and F are respectively `p` and `q`. If `n` independent trials are realized, then the probability to get `x` times of S is determined by:

`p(x)=P(X=x)=(n!)/(x!(n-x)!)\ p^x\ (1-p)^(n-x)`

(7)

which   `n!\ =nxx(n - 1)xx(n - 2)xx cdots xx3xx2xx1`

The probability distribution of this type of discrete random variable is denoted as binomial distribution.

We recognize that in formula (7), the component

  `((n),(x))=(n!)/(x!(n-x)!)`

is the combination of `n` taken `x`, which we already discussed in "Counting Rules".

Example 5 : In a box, there are 8 products of company A and 12 products of other companies. Take 7 products in succession with replacement from this box. What is the probability that there are 3 products of company A among 7 products ?

We can apply binomial distribution in this case with:

  `p` : probability that product taken belongs to company A: `p=8//(8+12)=0,4` ;
  `1-p=0,6`

Therefore :

  `p(3)=(7!)/(3!xx4!)xx0,4^3xx0,6^4=0,29`

 

Multinomial distribution is the expansion of binomial distribution, in which the number of outcome can be greater than 2.

Considering the random variable `X` has `k` outcomes A₁, A₂, ... , A_`i`, ... , A_`k` (`k>=2`), and the probability of outcome A_`i` is `p_i`. If we realized `n` independent trials, the probability that there are `x_1` times of getting A₁, `x_2` times of getting A₂, ... , `x_k` times of getting A_`k` can be determined by:

`P(X_1=x_1,\ X_2=x_2,\ ...\ ,X_k=x_k)=(n!)/(x_1!x_2!\ cdots\ x_k!)p_1^(x_1)p_2^(x_2)\ cdots\ p_k^(x_k)`

(8)

or

`P(X_1=x_1,\ X_2=x_2,\ ...\ ,X_k=x_k)=(n!)/(prod_(i=1)^k x_i!)prod_(i=1)^k p_i^(x_i)`

(9)

This distribution is denoted as multinomial distribution. We can see that binomial distribution is a special case of multinomial distribution with `k=2`.

The conditions to apply multinomial distribution are:

  `x_1+x_2+\ cdots\ +x_k=sum_(i=1)^k x_i=n`

and : `p_1+p_2+\ cdots\ +p_k=sum_(i=1)^k=1`

Example 6 : A box contains 8 products of company A, 2 products of company B, and 10 products of company C. Take 8 products in succession with replacement from this box. What is the probability that there are 3 products of company A, 1 product of company B and 4 products of company C ?

From this information, we recognize that :

• `p_"A"=8//20=0,4` ; `p_"B"=2//20=0,1` ; `p_"C"=10//20=0,5`
• `x_"A"=3` ; `x_"B"=1` ; `x_"C"=4` ; `n=8`
• All the conditions of multinomial distribution are satisfied.

Therefore :

  `P("E")=(8!)/(3!xx1!xx4!)xx0,4^3xx0,1^1xx0,5^4=0,112`