This is a discrete joint probability distribution. It is used when a particular trial is conducted ‘n’ number of times where each trial can have two or more outcomes. The binomial distribution is a special case of the multinomial distribution where each trial has exactly two outcomes.

For example, if we were to toss a coin 10 times then to calculate the probability of specific occurrences we would use the binomial distribution. But if we were to roll a die 10 times we would use the multinomial distribution to calculate the probability of a specific outcome occurring (since each trial can produce six outcomes in this case).

**Probability Mass Function of Multinomial Distribution**:

Suppose each trial can result in ‘k’ different outcomes. We shall assume that each trial is independent of the other (for example, the outcome of one coin toss does not influence the outcome of the successive coin toss). Let X_{i} (i=1,2,…,k) denote the number of times the i^{th} outcome occurs. Let p_{i} denote the probability that outcome ‘i’ occurs in a particular trial. Suppose n successive trials are conducted. The probability mass function is given as,

**Trinomial Distribution**:

The trinomial distribution is a special case of the multinomial distribution where n=3, that is, we have three possible outcomes in a particular Bernoulli trial.

For example, if a bag contains some balls of three different colours, then the number of balls selected of each type would jointly follow the trinomial distribution.

**Mean and Variance of Multinomial Distribution**:

**Example**: Suppose a box contains balls of three different colors- red, blue, and green. The box has 50% red balls,30% blue balls, and 20% green balls. Suppose 10 balls are chosen successively from the box. Calculate the probability that we choose 4 red balls,5 blue balls, and 1 green ball. Also, find the mean and variance for the number of red balls.

**Solution**: Let X_{1}, X_{2}, X_{3 }denote the number of red, blue, green balls chosen respectively. We are given p_{1}=0.5, p_{2}=0.3 and p_{3}=0.2.

Substituting all this in the above formula we get,

P(X_{1}=4, X_{2}=5, X_{3}=1) =(10! /4!*5!*1!)*(0.5)^{4}*(0.3)^{5}*(0.2)^{1}=**0.03827**

To calculate the mean and variance we use the formulae,

E(X_{1}) = np_{1} = 10*0.5 = **5**

V(X_{1}) = np_{1}q_{1}= 10*0.5*0.5 = **2.5**

**Marginal Distribution of X _{i}**:

If X_{1}, X_{2},…., X_{n} jointly follow the multinomial distribution then each of the random variables X_{i} follows the binomial distribution as the marginal distribution. This means that if we consider X_{i} as a random variable on its own it follows the binomial distribution.