@Diffusion -- at $\beta = 0$ it indeed blows up. I think that the statement still holds for $\beta > 0$ strictly. I modified the problem to avoid this edge case.

@Iosef: thank you, i fixed it

I realized that actually the bound should have the form $1 - e^{-\mu + \Delta}$. In your example, you would have $\Pr(B) = 1- \prod (1-p_i) \geq 1 - e^{-\mu}$, as desired. I apologize that I botched the original formula.

I adjusted the definition like you suggested. If $S_i$ is non-empty and does not depend on $i$, it would seem that $\Delta$ would be large. As I mentioned, I am interested in the case where $\mu \gg \Delta$.