Expected value of the sum of geometrically distributed variables Assume you have a printer which every minute attempts to print precisely one page. The probability of it being successful depends on the page to be printed. Where $p_i$ is the probability that page $i$ will be printed. The printer fails to print a page, it tries to print that same page again in the next minute—the probability of success being the same as before, independently of how many times it already tried to print that page. Once the printer managed to print the page, it moves on to the next page and stoically repeats its procedure (so never skips a page).
You send a job consisting of n pages to the printer and decide to come back and check on it after m minutes. What is the expected number of pages, when you try to print $n$ pages and check on it after $m$ minutes? 
I believe that each side is geomtrically distributed when viewed separately. But I don't know how to deal with the different probabilities of the single page. 
 A: Let $S$ be the random variable "number of printed pages at instant $m$".
Let $s_k:=P(S=k)$ (probability that exactly $k$ pages have been printed in this condition.
Let $q_i=1-p_i$.
Your issue can be modelized by the following "Finite State Machine" (https://en.wikipedia.org/wiki/Finite-state_machine#Finite_Markov_chain_processes) also called "automata" 

in connection with the so-called "Finite Markov Chains" that we will use here through the following formula (be conscious that there is a whole theory behind it !)
$$\underbrace{\begin{pmatrix}s_0\\s_1\\ \cdots \\s_{m-1}\\s_m\end{pmatrix}}_S=\underbrace{\left(\begin{array}{ccccc} q_1& 0&0&\cdots&0\\p_1&q_2&0&\cdots&0\\0&p_2&q_3&\cdots&0\\\cdots&&\cdots&&\cdots\\0&0&\cdots&p_{m-1}&q_{m}\end{array}\right)^m}_{M^m}\begin{pmatrix} 1\\0\\\cdots\\0\\0\end{pmatrix}\tag{1}$$
(as we have at most $m$ printed pages at this instant, no need to use a larger matrix).
For example, if $m=3$ (meaning that $m$ seconds have elapsed), we have (see definition of $s_k$ upwards), (1) will give :
$$\left\{\begin{array}{ccl}
 s_0&=&q_1^3\\
 s_1&=&q_1^2p_1 + q_1p_1q_2 + p_1q_2^2\\
 s_2&=&q_1p_1p_2 + p_1q_2p_2 + p_1p_2q_3\\
 s_3&=&p_1p_2p_3
\end{array}\right.\tag(2)$$
Explanation for $s_0$ : the probability that no page has been printed is the probability of 3 successive failures on page 1. 
Explanation for $s_1$ : 1 page exactly has been printed if we are in one of the three following (independent) cases : FFS, FSF or SFF (S= success, F= failure), etc...

Therefore, 
$$E(S) = 0 \times s_0 + 1 \times s_1 + 2 \times s_2 + \cdots + m \times s_m$$

Remarks : 
1) If indices are erased in (2), one gets... a binomial distribution.
2) Formula (1) could be expressed under a different form : $S$ is the first column of $M^n$.
