EXAMPLE 5.9

A binomial experiment consists of $n$ dichotomous (success-failure), homogenous (constant success probability) independent trials. Now consider a trinomial experiment in which each of the $n$ trials can result in one of three possible outcomes. For example, each successive customer at a store might pay with cash, a credit card, or a debit card. The trials are assumed independent. Let $p_1=P$ (trial results in a type 1 outcome) and define $p_2$ and $p_3$ analogously for type 2 and type 3 outcomes. The random variables of interest here are $X_i=$ the number of trials that result in a type $i$ outcome for $i=1,2,3$.

In $n=10$ trials, the probability that the first five are type 1 outcomes, the next three are type 2, and the last two are type 3-that is, the probability of the experimental outcome 1111122233-is $p_1^5\cdot p_2^3\cdot p_3^2$. This is also the probability of the outcome 1122311123, and in fact the probability of any outcome that has exactly five 1 ’s, three 2’s, and two 3’s. Now to determine the probability $P(X_1=5,X_2=3$, and $X_3=2$ ), we have to count the number of outcomes that have exactly five 1 ’s, three 2 ’s, and two 3’s. First, there are $\left(\begin{array}{c}10\\ 5\end{array}\right)$ ways to choose five of the trials to be the type 1 outcomes. Now from the remaining five trials, we choose three to be the type 2 outcomes, which can be done in $\left(\begin{array}{l}5\\ 3\end{array}\right)$ ways. This determines the remaining two trials, which consist of type 3 outcomes. So the total number of ways of choosing five 1 ‘s, three 2’s, and two 3’s is

$$\left(\begin{array}{c}10\\ 5\end{array}\right)\cdot \left(\begin{array}{l}5\\ 3\end{array}\right)=\frac{10!}{5!5!}\cdot \frac{5!}{3!2!}=\frac{10!}{5!3!2!}=2520$$

Thus we see that $P\left(X_1=5,X_2=3,X_3=2\right)=2520p_1^5\cdot p_2^3\cdot p_3^2$. Generalizing this to $n$ trials gives

$$p\left(x_1,x_2,x_3\right)=P\left(X_1=x_1,X_2=x_2,X_3=x_3\right)=\frac{n!}{x_1!x_2!x_3!}{p}_1^{x_1}{p}_2^{x_2}{p}_3^{x_3}$$

for $x_1=0,1,2,\dots ;x_2=0,1,2,\dots ;x_3=0,1,2,\dots$ such that $x_1+x_2+x_3=n$. Notice that whereas there are three random variables here, the third variable $X_3$ is actually redundant. For example, in the case $n=10$, having $X_1=5$ and $X_2=3$ implies that $X_3=2$ (just as in a binomial experiment there are actually two rv’s-the number of successes and number of failures - but the latter is redundant).

As a specific example, the genetic allele of a pea section can be either AA, Aa, or aa. A simple genetic model specifies $P\left(\mathrm{AA}\right)=.25,P\left(\mathrm{Aa}\right)=.50$, and $P\left(\mathrm{aa}\right)=.25$. If the alleles of 10 independently obtained sections are determined, the probability that exactly five of these are $\mathrm{Aa}$ and two are $\mathrm{AA}$ is

$$p\left(2,5,3\right)=\frac{10!}{2!5!3!}{\left(.25\right)}^2{\left(.50\right)}^5{\left(.25\right)}^3=0.769$$

A natural extension of the trinomial scenario is an experiment consisting of $n$ independent and identical trials, in which each trial can result in any one of $r$ possible outcomes. Let $p_i=P$ (outcome $i$ on any particular trial), and define random variables by $X_i=$ the number of trials resulting in outcome $i\left(i=1,\dots ,r\right)$. This is called a multinomial experiment, and the joint pmf of $X_1,\dots ,X_r$ is called the multinomial distribution. An argument analogous to the one used to derive the trinomial pmf gives the multinomial pmf as

$$\begin{array}{rl}p\left(x_1,\dots ,x_r\right)& =\{\begin{array}{ll}\frac{n!}{\left(x_1!\right)\left(x_2!\right)\cdots \left(x_r!\right)}\cdot p_1^{x_1}\cdots p_r^{x_r}& \text{ if }{x}_i=0,1,2,\dots ;x_1+\cdots +x_r=n\\ 0& \text{ otherwise }\end{array}\end{array}$$