[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)\mathbb P(B)[/tex].
So,
[tex]\mathbb P(A\cup B)=0.4+0.1+0.4\times0.1=0.54[/tex]
