The multinomial distribution or probability is just a generalization of the binomial distribution or probability. Binomial distribution is used in cases where we have only two possible outcomes. But we use multinomial distribution in cases where there are more than two outcomes.

**For Example:** the experiment of flipping a coin has only two outcomes while experiment of throwing a dice has six outcomes. So in case of large number of trials of throwing dice experiment we would require the use of multinomial probability and not binomial probability.

A multinomial experiment possesses following properties:

A multinomial distribution is the distribution of probability in which the outcomes are coming from a multinomial experiment.

If
we made 'n' number of trials in an experiment out of which 'k' 'types
of outcomes' probability can then be determined by using the formula for
multinomial probability given below:

P= $\frac{n!}{(n_1!)(n_2!)….(n_k!)}$ . $(p_1^{n_1} p_2^{n_2} …. p_k^{n_k})$

Here, P is the multinomial probability for the given experiment.

'n' is the number of total trials made in the experiment.

$n_1$, is the number of occurrences of the first outcome

$n_2$ is the number of occurrences of the second outcome …, and so on

$n_k$ is the number of occurrences of nth outcome.

And similarly $p_1$, is the probability of occurrence of the first outcome

$p_2$ is the probability of occurrence of the second outcome …, and so on

$p_k$ is the probability of occurrence of nth outcome

The
binomial probability is a special case of this multinomial probability
only. When we have k = 2, then multinomial probability changes itself to
binomial probability.

Few examples based on multinomial probability are as follow:

**Example 1:** An experiment of drawing a random card from an ordinary playing cards deck is done with replacing it back. This was done ten times. Find the probability of getting 2 spades, 3 diamond, 3 club and 2 hearts.

We use the multinomial probability here to solve this problem.

In this case, the number of trials, n = 10

Now since four types of outputs are there means k = 4 with $n_1$ = 2, $n_2$ = 3, $n_3$ = 3, $n_4$ = 2

Now the probability of drawing a spade, diamond, club or a heart is $\frac{13}{52}$ = 0.25 for each one of them. Thus we get that $p_1$ = 0.25, $p_2$ = 0.25 = $p_3$ = $p_4$.

Now we substitute the values in the multinomial probability formula and we get,

P = $[\frac{n!}{(n_1! . n_2! . n_3! . n_4!)}]$ . $p_1^{n_1}$ . $p_2^{n_2}$ . $p_3^{n_3}$ . $p_4^{n_4}$

$\rightarrow$ P = $[\frac{10!}{(2! . 3! . 3! . 2!)}]$ . $(0.25)^2$ . $(0.25)^3$ . $(0.25)^3$ . $(0.25)^2$

$\rightarrow$ P = 0.024

**Example 2:** Of the 10 bits received, what is the probability that 5 are excellent, 2 are good and 2 are fair and 1 is poor? Assume that the classification of individual bits are independent events and that the probabilities of A, B, C and D are 40%, 20%, 5% and 1% respectively.

**Solution:**

Total number of bits received, n = 10

Also $n_1$ = 5, $n_2$ = 2, $n_3$ = 2, $n_4$ = 1

The probabilities of A, B, C and D are 40%, 20%, 5% and 1% respectively i.e.

$p_1$ = 0.4, $p_2$ = 0.2, $p_3$ = 0.05 and $p_4$ = 0.01

The requested probability is:

P = $\frac{10!}{5!2!2!1!}$ $\times$ ((0.4)$^{5}$(0.2)$^{2}$(0.05)$^{2}$(0.01)$^{1})$

= $\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!2!2!1!}$

((0.4)$^{5}$(0.2)$^{2}$(0.05)$^{2}$(0.01)$^{1})$

= 7560 (1024 $\times$ 10$^{-11}$ )

= 7741440 $\times$ 10$^{-11}$

Total number of bits received, n = 10

Also $n_1$ = 5, $n_2$ = 2, $n_3$ = 2, $n_4$ = 1

The probabilities of A, B, C and D are 40%, 20%, 5% and 1% respectively i.e.

$p_1$ = 0.4, $p_2$ = 0.2, $p_3$ = 0.05 and $p_4$ = 0.01

The requested probability is:

P = $\frac{10!}{5!2!2!1!}$ $\times$ ((0.4)$^{5}$(0.2)$^{2}$(0.05)$^{2}$(0.01)$^{1})$

= $\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5!2!2!1!}$

((0.4)$^{5}$(0.2)$^{2}$(0.05)$^{2}$(0.01)$^{1})$

= 7560 (1024 $\times$ 10$^{-11}$ )

= 7741440 $\times$ 10$^{-11}$