probability of exactly one out of N events occuring I have N events.  Each "i" event has probability $P_i$.  What is the probability of $n$ events occuring?  I have seen this answered for two and three events, but not for an arbitrary N.  In principle, I guess the answer for one event would be
$\sum_i P_i \prod_{j\ne i}(1-P_j)$
but to meaningfully solve the problem I am working I need a much more compact way to express this, and I am not seeing it.
If there is no general answer to this, then specifically, the probability for event $i$ is $P_i=1/i$, if that is needed.
 A: Assuming the events are independent, you are looking for the coefficient of $x^n$ in the expansion of $$\displaystyle\prod_{i=1}^N \left(1+(x-1)P_i\right)$$ or in your particular example with $P_i=\frac1i$  the coefficient of $x^n$ in the expansion of $$\displaystyle\prod_{i=1}^N \frac{x+i-1}{i}.$$
The answer is in fact $\dfrac{s_1(N,n)}{N!}$ where $s_1(N,n)$ represents a Stirling number of the first kind and  $s_1(N,n) = s_1(N-1,n)+ (N-1) s_1(N-1,n-1)$. 
A: Let's see what the probability for a specific $n$ event. Suppose $n=\{k_1,k_2, \ldots k_n\}$ happens and $N \setminus \{n\}$ don't. Exactly as you guessed we have
$$\prod_{k\in n} P_{k} \prod_ { i \in N \setminus n} (1- P_i)$$
Now we need all possible cases of this...So we sum over all permutations of $n$. i.e.
$$\mathbb{P}( \text{only $n$ events happen}) = \sum_{\sigma \in S_n}\prod_{k\in \sigma} P_{k} \prod_ { i \in N \setminus \{\sigma\}} (1- P_i)$$
where $S_n$ is the symmetric group on $n$ letters. Now given $P_i = 1/i$, we see
$$\mathbb{P}( \text{only $n$ events happen}) = \sum_{\sigma \in S_n}\prod_{k\in \sigma}  \prod_ { i \in N \setminus \sigma} \frac{i-1}{ik}$$
I'm not sure how to simplify
A: First, we rewrite your (correct) expression as
$$\sum_{i=1}^{N}P_i\prod_{{i\neq j}} (1-P_j)=\sum_{i=1}^{N}\frac{P_i}{1-P_i}\prod_{j=1}^{n}(1-P_j),$$
and plug in $P_i=\frac{1}{i}$ to find that, in our case, when $N>1$, we are interested in
$$\sum_{i=2}^{N}\frac{P_i}{1-P_i}\prod_{j=2}^{N}(1-P_j)=\frac{1}{N}\sum_{i=1}^{N-1}\frac{1}{i}$$
