Probability Distribution with different probabilities Suppose there are 9 events, that have a probability of 10%, 20%, 30%, ..., 90% of being a success.
How would I find the probability of exactly n number of these events succeeding?
For n = 1, I'm thinking it is something like
(first succeeds) -> (1-0.9) * 0.8 * 0.7 * 0.6 ... 0.1
(second succeeds) -> 0.9 * (1-0.8) * 0.7 * 0.6 ... 0.1
(...)
(last)

and then adding them up.
The probabilities might not always be an an Arithmetic Progression, hoping to find a solution that doesn't depend on that.
 A: The generating function for the probabilities $a_n$ for $n$ successes is
$$\sum_{n=0}^9 a_n t^n=\prod_{k=1}^9(p_k t+(1-p_k))$$
where $p_k$ is the probability of the $k$-th event occuring, in this
example $p_k=k/10$. Maybe in this case the product won't simplify
so nicely, but the method is quite general.
A: Take the events in order.  After the first, you have 0.1 chance of 1 success and 0.9 chance of 0 successes.  Then add on the second.  The chance of one success after two events is 0.2*the chance of failure on the first + 0.8*the chance of success on the first.  A spreadsheet is your friend for this.  After all the events, you will have a column with the chance of each possible number of successes.  Adding to see if you still get 1 is a good check that your equations are right.  If you use the absolute/relative reference codes correctly, you can copy down and right most of the equations.
A: Let the probability that the $i$th event occurs be $p_i = i/10$ and let the probability that exactly $n$ events occur be $P(n).$
So the probability that no events occur is $P(0) = \prod_{i=1}^9 (1-p_i) = 3.6288 \times 10^{-4}.$
Let $S= \lbrace 1,2,3,\ldots,9 \rbrace $ then for $n \ge 1$ we have
$$P(n) = P(0) \times \sum_{i_1,i_2,\ldots,i_n \in S, \quad i_k \textrm { distinct} }
\frac{p_{i_1} p_{i_2} \ldots p_{i_n} } {(1-p_{i_1})(1-p_{i_2}) \ldots (1-p_{i_n}) },$$
where there are ${9 \choose n}$ terms in the summation.
