Law of Total Probability
Theorem
Given a set of events \(\{B_i\}_{i=1}^n\) which partition the sample space, and some event \(A\), we have that:
\[ P(A) = \sum_{i=1}^n P(A \cap B_i) = \sum_{i=1}^n P(A \mid B_i)P(B_i).\]
This can be intuitively thought of as the partition of the whole space yielding a partition of \(A\):
and then the probability of \(A\) calculated by summing the part in each set within the partition
Proof
The result follows very simply from the ability to split the measure across a disjoint union:
\[\begin{align*}
P(A) &= P\left(\dot\bigcup_{i=1}^n (A \cap B_i)\right) \\
&= \sum_{i=1}^n P(A \cap B_i),
\end{align*}\]
and hence, the result we must prove that \(A \cap B_i\) and \(A \cap B_j\) are disjoint for \(i \neq j\).
This is because:
\[\begin{align*}
(A \cap B_i) \cap (A \cap B_j) &= (A \cap A) \cap (B_i \cap B_j) \\
&= A \cap \varnothing \\
&= \varnothing \\
\end{align*}\]
where \(B_i \cap B_j = \varnothing\) from the assumption that \(\{B_i\}_{i=1}^n\) partitions the sample space.