Binomial, Poisson, and Multinomial Distributions
Learning Objectives
- Define multinomial outcomes
- Compute probabilities using the multinomial distribution
\(\begin{align*}
\mathrm{p}=\frac{\mathrm{n} !}{\left(\mathrm{n}_{1} !\right)\left(\mathrm{n}_{2} !\right)\left(\mathrm{n}_{3} !\right)} \mathrm{p}_{1}^{\mathrm{n}_{1}} \mathrm{p}_{2}^{\mathrm{n}_{2}} \mathrm{p}_{3}^{\mathrm{n}_{3}}
\end{align*}\)
where
\(\mathrm{p}\) is the probability,
\(\mathrm{n}\) is the total number of events
\(\mathrm{n}_{1}\) is the number of times outcome 1 occurs,
\(\mathrm{n}_{2}\) is the number of times outcome 2 occurs,
\(\mathrm{n}_{3}\) is the number of times outcome 3 occurs,
\(\mathrm{p}_{1}\) is the probability of outcome 1
\(\mathrm{p}_{2}\) is the probability of outcome 2 , and
\(\mathrm{p}_{3}\) is the probability of outcome 3.
For the chess example,
\(\mathrm{n}=12\) (12 games are played),
\(\mathrm{n}_{1}=7\) (number won by Player A),
\(\mathrm{n}_{2}=2\) (number won by Player B),
\(\mathrm{n}_{3}=3\) (the number drawn),
\(\mathrm{p}_{1}=0.40\) (probability Player A wins)
\(\mathrm{p}_{2}=0.35\) (probability Player B wins)
\(\mathrm{p}_{3}=0.25\) (probability of a draw)
\(\mathrm{p}=\frac{12 !}{(7 !)(2 !)(3 !)} \cdot 40^{7} \cdot 35^{2} \cdot 25^{3}=0.0248\)
The formula for \(\mathrm{k}\) outcomes is
\(\begin{align*}
\mathrm{p}=\frac{\mathrm{n} !}{\left(\mathrm{n}_{1} !\right)\left(\mathrm{n}_{2} !\right) \ldots\left(\mathrm{n}_{\mathrm{k}} !\right)} \mathrm{p}_{1}^{\mathrm{n} 1} \mathrm{p}_{2}^{\mathrm{n}_{2}} \ldots \mathrm{p}_{\mathrm{k}}^{\mathrm{n}}
\end{align*}\)
Note that the binomial distribution is a special case of the multinomial when \(\mathrm{k}=\) \(2\).