Law of total probability

Given event $A$ where $0<P(A)<1$, then for any event $B$ we have

$$P(B)=P(B\mid A)\cdot P(A)+P(B\mid A^{\complement })\cdot P(A^{\complement })$$

The general case holds for events $A_1,A_2,...,A_n$ such that they partition the sample space $S$. That is, they are all pairwise disjoint:

$$P(B)=\sum _{i=1}^{n}P(B\mid A_i)\cdot P(A_i)$$

We can think of the “$\cdot P(A_i)$” part as multiplying the result by the “weight” of the case that event $A_i$ occurs.

Intuitively, if $P(A_i)>P(A_j)$, then when we sum them for $P(B)$, the “case” of $P(A_i)$ should matter more in the overall sum for $P(A)$ than $P(A_j)$.