A and B must be disjoint. You have, by the inclusion/exclusion principle, that
\(\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)\)
where the last term is required to prevent double counting. When \(A\) and \(B\) are disjoint, you have \(A\cap B=\emptyset\) and \(\mathb P(A\cap B)=0\), leaving you with
\(\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)\)
