I got stuck with this:
P(A|C) = $\sum_{B}$ (A, B|C)
(B should be beneath the sum sign)
I understand marginal probability with two variables. But when it is three or more variables, I just could not make sense of it. What if A intersects C, but it does not intersect with the intersection of B and C? Why sum of B works? Can someone walk me through this? Thanks!
You should be aware of the definition of conditional probability. $\mathsf P(A\mid C)=\mathsf P(A,C)/\mathsf P(C)$
For three events, same again: $\mathsf P(A, B\mid C)=\mathsf P(A,B,C)/\mathsf P(C)$
So just apply that and the Law of Total Probability.
$$\begin{align}\mathsf P(A\mid C) ~=~&\dfrac{\mathsf P(A,C)}{\mathsf P(C)} \\[1ex]=~& \dfrac{\sum_{B_k}\mathsf P(A,B_k,C)}{\mathsf P(C)} \\[1ex]=~& \sum_{B_k}\mathsf P(A, B_k\mid C)\end{align}$$
That is all.
As to why the sum works: If $(B_k)$ is a sequence of disjoint events partitioning the sample space -- well, in this case, at least exhaustively covering $C$ -- then: $$\mathsf P(A\cap C) ~=~ \mathsf P(\bigcup_{B_k} A\cap B_k\cap C)~=~\sum_{B_k}\mathsf P(A\cap B_k\cap C)$$