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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.