For [tex]A[/tex] and [tex]B[/tex] to be independent, it must be the case that
[tex]\mathbb P(A\cap B)=\mathbb P(A)\times\mathbb P(B)[/tex]
You're given
[tex]\mathbb P(A\cap B)=\dfrac16[/tex]
while
[tex]\mathbb P(A)\times\mathbb P(B)=\dfrac14\times\dfrac23=\dfrac16[/tex]
So yes, the two events are independent.
