Recall that
[tex]\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)[/tex]
Since [tex]A[/tex] and [tex]B[/tex] are independent, we have [tex]\mathbb P(A\cap B)=\mathbb P(A)\cdot\mathbb P(B)[/tex]. So
[tex]\mathbb P(A\cup B)=0.2+0.4-0.2\cdot0.4=0.52[/tex]
