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