Let $B \subseteq \biguplus^{\infty}_{n=1} A_n$, show that $\mathbb{P}(B)=\sum^{\infty}_{n=1} \mathbb{P}(A_n) \mathbb{P}(B|A_n)$

Question

Question:

Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space with events $A,B\in\mathcal{A}$. Now, let $B \subseteq \biguplus^{\infty}_{n=1} A_n$, where $A_n \in \mathcal{A}$ for each $ n \in \mathbb{N}$.

Show that: $\mathbb{P}(B)=\sum^{\infty}_{n=1} \mathbb{P}(A_n) \mathbb{P}(B|A_n)$

My attempt so far was to restructure the right-hand side to $\sum^{\infty}_{n=1}\mathbb{P}(A_n)\frac{\mathbb{P}(B \cap A_n)}{\mathbb{P}(A_n)} = \sum^{\infty}_{n=1}\mathbb{P}(B \cap A_n)$ but I don't know if I'm even on the right track, so I'm pretty much stuck at this point.

Answer 1

Hint: if $\biguplus$ means disjoint union, $$B = B\cap\left(\biguplus_{n=1}^\infty A_n\right) = \biguplus_{n=1}^\infty(B\cap A_n). $$