sum of two proportions I have 10 sets of 100 marbles.  I have two processes that pick marbles.  One of them usually picks between 2 and 15% of the marbles in the set.  Another picks between 0 and 90% of the marbles in the set, independently of the first process, so I guess with replacement.  For each set I know how many marbles were picked by each process, but not the total number picked together -- that's what i want to know.
My intuition is that when the proportions are small, adding them is fine, but not when the sum gets close to 100%.  Also, for my use case, saying 100% have been picked is very impactful so I don't want to say 100% have been picked unless my confidence in that is really high. 
Given that there are only 100 marbles it seems unlikely that in the case that one process picks 11% and the other picks 90% that actually 100% (101%?) of the marbles have been picked, but that's naively how I know to do it.  What's the better way, that takes into account the limited number of marbles?  Thank you!
 A: Here's the way I would approach it:
If $A$ is the number of marbles chosen by the first process, $B$ is the number of marbles chosen by the second process, and $k$ is the number of marbles that were chose by both processes, then the total number of marbles chosen is $A + B - k$. 
I'm assuming we know $A$ and $B$, and what we're trying to estimate is $A + B - k$.  In other words, $k$ is what adds randomness to the process.
In this scenario, $k$ follows a hypergeometric distribution, with parameters: $N = 100, K = A,$ and $n = B$, and the expected value of $k$ is $\frac{nK}{N}$.
For example, say the first process picks 10 of the marbles and the second process picks 80 of the marbles.  Then the number of marbles chosen by both processes is hypergeometric with parameters $N = 100, K = 10,$ and $n = 80$, and we would expect that $\frac{80\cdot 10}{100} = 8$ marbles were chosen by both processes. 
Then the expected number of total marbles picked is $10 + 80 - 8 = 92$.
The intuition is that if you picked 10 marbles in the first process, then when you pick 80 (80% of the total) during the second process, you expect to get 80% of those first 10 marbles, which is 8, as overlap.
