= Solution
Let $(B_j)$ be a finite or countable <partition of a set> of the <sample space> into disjoint <events>, with $\mathbb P(B_j)>0$. For any <event> $A$, the <law of total probability> is
$$
\boxed{\mathbb P(A)=\sum_j\mathbb P(A\mid B_j)\mathbb P(B_j).}
$$
To prove it, the intersections $A\cap B_j$ are disjoint and have union $A$. <Countable additivity> and the definition of <conditional probability> therefore give
$$
\mathbb P(A)=\sum_j\mathbb P(A\cap B_j)=\sum_j\mathbb P(A\mid B_j)\mathbb P(B_j).
$$
It is enough that the union of the partition <events> has <probability> one. Parts of <probability> zero contribute zero in the intersection formula and can be omitted before writing <conditional probabilities>; conditioning by division on a zero-<probability> part would be undefined.
Back to article page
