Find a probability of $n$ event happening from $m$ types

Question

The question is: to find a sum $$ S=\sum\limits_{n_1+n_2+\ldots+n_m = n,\ n_i=0,1,\ldots,n} p_1^{n_1}p_2^{n_2}\cdots p_m^{n_m}, $$ where $p_i\in[0,1]$.

UPDATE. This issue has no probabalistic meaning, sorry for misleading you.

Answer 1

It seems the following.

Without loss of generality we may assume that all numbers $p_i$ are mutually different, because in the opposite case we simply decrease $m$. Denote $S$ as $S(n;p_1,\dots,p_m)$.

If $m=1$ then $S(n;p_1)=p_1^n$.

If $m=2$ then $S(n;p_1,p_2)=\frac{p_1^{n+1}- p_2^{n+1}}{p_1-p_2}$.

Let $m\ge 3$. Then

$$S(n;p_1,\dots,p_m)=\sum_{k=0}^n \sum_{n_{m-1}+n_m=k} S(n-k;p_1,\dots,p_{m-2})p_{m-1}^{n_{m-1}} p_m^{n_m}=$$ $$\sum_{k=0}^n \sum_{n_{m-1}+n_m=k} S(n-k;p_1,\dots,p_{m-2}) \frac{p_m^{k+1} - p_{m-1}^{k+1}}{p_m -p_{m-1}}=$$ $$\frac{1}{p_m -p_{m-1}}\left(S(n+1;p_1,p_2,\dots,p_{m-2},p_m)- S(n+1;p_1,\dots,p_{m-1}) \right).$$

then by recursion we obtain

$$S(n;p_1,\dots,p_m)=\sum_{i=1}^m p_i^{n+m-1}\prod_{j\ne i} (p_i-p_j)^{-1}.$$

Comparing this expression with Lagrange interpolation formula, we can conclude that $S(n;p_1,\dots,p_m)$ is equal to the highest coefficient of an unique $(m-1)$-degree polynomial $P$ such that $P(x)=x^{n+m+2}$ for each $x\in\{p_1,\dots,p_m\}$. :-)