Statistics - Multinomial Distribution

A multinomial experiment is a statistical experiment and it consists of n repeated trials. Each trial has a discrete number of possible outcomes. On any given trial, the probability that a particular outcome will occur is constant.

Formula

${P_r = \frac{n!}{(n_1!)(n_2!)...(n_x!)} {P_1}^{n_1}{P_2}^{n_2}...{P_x}^{n_x}}$

Where −

${n}$ = number of events
${n_1}$ = number of outcomes, event 1
${n_2}$ = number of outcomes, event 2
${n_x}$ = number of outcomes, event x
${P_1}$ = probability that event 1 happens
${P_2}$ = probability that event 2 happens
${P_x}$ = probability that event x happens

Example

Problem Statement:

Three card players play a series of matches. The probability that player A will win any game is 20%, the probability that player B will win is 30%, and the probability player C will win is 50%. If they play 6 games, what is the probability that player A will win 1 game, player B will win 2 games, and player C will win 3?

Solution:

Given:

${n}$ = 12 (6 games total)
${n_1}$ = 1 (Player A wins)
${n_2}$ = 2 (Player B wins)
${n_3}$ = 3 (Player C wins)
${P_1}$ = 0.20 (probability that Player A wins)
${P_1}$ = 0.30 (probability that Player B wins)
${P_1}$ = 0.50 (probability that Player C wins)

Putting the values into the formula, we get:

${ P_r = \frac{n!}{(n_1!)(n_2!)...(n_x!)} {P_1}^{n_1}{P_2}^{n_2}...{P_x}^{n_x} , \\[7pt] \ P_r(A=1, B=2, C=3)= \frac{6!}{1!2!3!}(0.2^1)(0.3^2)(0.5^3) , \\[7pt] \ = 0.135 }$

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