In one experiment we consider three events $A_1$, $A_2$ and $A_3$, with probabilities $p_1$,$p_2$ and $p_3$, such that $p_1+p_2+p_3=1 $
If we consider a sequence of $n$ independent trials of the experiment, the trinomial distribution arises when there are three possible outcomes. We can consider three r.v $X_i$ equal to the number of times that $A_i(i=1,2,3)$ occurs.
$X_1+X_2+X_3=n$.
$n_i$ is the value that the r.v $X_i$ takes.
$P(n_1,n_2,n_3)=\frac{n!}{n_1!n_2!n_3!}p_1^{n_1}p_2^{n_2}p_3^{n_3}$
My question arises when you find the marginal probability of $X_1$
$$P(n_1)=\sum_{n_2}P(n_1,n_2)=$$ $$\binom{n}{n_1}p_1^{n_1}\sum_{n_2=0}^{n-n_1}\binom{n-n_1}{n_2}p_2^{n_2}p_3^{n-n_1-n_2}=$$ $$\binom{n}{n_1}p_1^{n_1}(p_2+p_3)^{n-n_1}=\binom{n}{n_1}p_1^{n_1}(1-p_1)^{n-n_1} $$
In the proof above why
$$\sum_{n_2=0}^{n-n_1}\binom{n-n_1}{n_2}p_2^{n_2}p_3^{n-n_1-n_2}$$equals $$(p_2+p_3)^{n-n_1} ?
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Also in the sum $$\sum_{n_2=0}^{n-n_1}$$ the subindex $n_2$ starts at $0$. Is it right to consider the case that the event $A_2$ doesn't occur?
Best Answer
This is an application of the binomial theorem,
$$ (x+y)^m=\sum_{k=0}^m\binom mkx^ky^{m-k}\;, $$
with $x=p_2$, $y=p_3$ and $m=n-n_1$.
I'm afraid I don't understand why you doubt that it's right to consider the case that the event $A_2$ doesn't occur.