Multinomial Distribution The binomial distribution applies only to cases involving two types of outcomes, whereas the multinomial distribution involves more than two categories. Suppose we have three types of mutually exclusive outcomes denoted by A, B, and C. Let\(P\left( A \right) = {p_1}\),\(P\left( B \right) = {p_2}\)and\(P\left( C \right) = {p_3}\). In n independent trials, the probability of\({x_1}\)outcomes of type A,\({x_2}\)outcomes of type B, and\({x_3}\)outcomes of type C is given by

\[\frac{{n!}}{{\left( {{x_1}} \right)!\left( {{x_2}} \right)!\left( {{x_3}} \right)!}}{p_1}^{{x_1}} \times {p_2}^{{x_2}} \times {p_3}^{{x_3}}\]

A roulette wheel in the Venetian casino in Las Vegas has 18 red slots, 18 black slots, and 2 green slots. If roulette is played 15 times, find the probability of getting 7 red outcomes, 6 black outcomes, and 2 green outcomes.

A roulette wheel was played 15 times. It has 18 red slots, 18 black slots, and 2 green slots.

If a procedure involves more than two types of independent outcomes of more than 2 types of categories, then such a procedure can be modeledusing a multinomial distribution.

The number of categories is equal to the number of colorsof slots = 3.

The total number of trials (n) is equal to 38.

Let\({x_1}\)represent the number of times red slots appear.

Let\({x_2}\)represent the number of times black slots appear.

Let\({x_3}\)represent the number of times green slots appear.

Here,

\(\begin{array}{l}{x_1} = 7\\{x_2} = 6\\{x_3} = 2\end{array}\)

Now, the total number of slots in the wheel is equal to:

\(18 + 18 + 2 = 38\)

The probability of getting a red slot appear is equal to:

\[\begin{array}{c}{p_1} = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{red}}\;{\rm{slots}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{slots}}}}\\ = \frac{{18}}{{38}}\\ = 0.4737\end{array}\]

The probability of getting a black slot appear is equal to:

\[\begin{array}{c}{p_2} = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{black}}\;{\rm{slots}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{slots}}}}\\ = \frac{{18}}{{38}}\\ = 0.4737\end{array}\]

The probability of getting a green slot appear is equal to:

\[\begin{array}{c}{p_3} = \frac{{{\rm{Number}}\;{\rm{of}}\;{\rm{green}}\;{\rm{slots}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{slots}}}}\\ = \frac{2}{{38}}\\ = 0.0526\end{array}\]

The multinomial probability used is as follows:

\[P\left( {{X_1} = {x_1},{X_2} = {x_2},{X_3} = {x_3}} \right) = \frac{{n!}}{{\left( {{x_1}} \right)!\left( {{x_2}} \right)!\left( {{x_3}} \right)!}}{p_1}^{{x_1}} \times {p_2}^{{x_2}} \times {p_3}^{{x_3}}\]

Thus, the probability of getting 7 red outcomes, 6 black outcomes, and 2 green outcomes is computed below:

\[\begin{array}{c}P\left( {{\rm{Red}} = 7,{\rm{Black}} = 6,{\rm{Green}} = 2} \right) = \frac{{15!}}{{\left( 7 \right)!\left( 6 \right)!\left( 2 \right)!}} \times {\left( {0.4737} \right)^7} \times {\left( {0.4737} \right)^6} \times {\left( {0.0526} \right)^2}\\ = 0.0301\end{array}\]

Therefore, the probability of getting 7 red outcomes, 6 black outcomes, and 2 green outcomes is equal to 0.0301.