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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 −
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${n}$ = number of events
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${n_1}$ = number of outcomes, event 1
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${n_2}$ = number of outcomes, event 2
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${n_x}$ = number of outcomes, event x
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${P_1}$ = probability that event 1 happens
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${P_2}$ = probability that event 2 happens
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${P_x}$ = 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:
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${n}$ = 12 (6 games total)
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${n_1}$ = 1 (Player A wins)
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${n_2}$ = 2 (Player B wins)
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${n_3}$ = 3 (Player C wins)
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${P_1}$ = 0.20 (probability that Player A wins)
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${P_1}$ = 0.30 (probability that Player B wins)
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${P_1}$ = 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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