Heuristic analysis of Bloom filters
A bit is set to 1 if it has been hit. It has $k|S|$ chances of being hit, and each time it is hit with probability $1/n$. Using a union bound, the probability that a bit is hit is at most $k|S|/n$.

We can improve on this bound by calculating instead the probability that a bit is 0, that is, that it is not hit. For a bit not to be hit, it has to be missed each time. Since it has $k|S|$ opportunities to be hit, and each time it is missed with probability $1-1/n$, in total a bit is missed with probability $(1-1/n)^{k|S|}$. Thus a bit is not missed with probability $1-(1-1/n)^{k|S|}$.