[tex]\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)[/tex]
If [tex]A[/tex] and [tex]B[/tex] are independent, then [tex]\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)[/tex], so under this assumption
[tex]\mathbb P(A)\mathbb P(B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cup B)[/tex]
The only way for [tex]\mathbb P(A\cup B)[/tex] to be equal to [tex]\mathbb P(A)\mathbb P(B)[/tex] is if both [tex]\mathbb P(A)[/tex] and [tex]\mathbb P(B)[/tex] are zero.
