The Multinomial Distribution

Generalizes the trinomial distribution to any number of outcomes per trial. Students compute joint probabilities using the multinomial PMF for direct word-problem scenarios involving independent trials with k possible outcomes.

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The Multinomial Distribution

The multinomial distribution generalizes the trinomial distribution to any number of outcomes k2.k \geq 2.

Suppose we perform nn independent trials, where each trial results in exactly one of kk possible outcomes with respective probabilities p1,p2,,pkp_1, p_2, \dots, p_k satisfying

p1+p2++pk=1.p_1 + p_2 + \cdots + p_k = 1.

Let XiX_i denote the number of trials that produce outcome i.i. Then for any nonnegative integers x1,x2,,xkx_1, x_2, \dots, x_k with x1+x2++xk=n,x_1 + x_2 + \cdots + x_k = n,

P(X1=x1,X2=x2,,Xk=xk)=n!x1!x2!xk!p1x1p2x2pkxk.P(X_1=x_1,\, X_2=x_2,\, \dots,\, X_k=x_k) = \dfrac{n!}{x_1!\, x_2! \cdots x_k!}\, p_1^{x_1}\, p_2^{x_2} \cdots p_k^{x_k}.

The coefficient n!x1!x2!xk!\dfrac{n!}{x_1!\, x_2! \cdots x_k!} is called the multinomial coefficient.

For example, roll a fair 44-sided die 55 times. The probability of getting two 11s, one 2,2, one 3,3, and one 44 is

P(X1=2,X2=1,X3=1,X4=1)=5!2!1!1!1!(14)2 ⁣(14) ⁣(14) ⁣(14)=6011024=15256.\begin{align*} P(X_1{=}2,X_2{=}1,X_3{=}1,X_4{=}1) &= \dfrac{5!}{2!\,1!\,1!\,1!}\left(\dfrac{1}{4}\right)^2\!\left(\dfrac{1}{4}\right)\!\left(\dfrac{1}{4}\right)\!\left(\dfrac{1}{4}\right) \\[3pt] &= 60 \cdot \dfrac{1}{1024} \\[3pt] &= \dfrac{15}{256}. \end{align*}
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