# Sum in the marginal distribution of the trinomial

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 $$(p_2+p_3)^{n-n_1} ?$$

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?

$$(x+y)^m=\sum_{k=0}^m\binom mkx^ky^{m-k}\;,$$
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.