For this question, we can draw a table to see the possible outcomes.
A total of 7 can be obtained by 6 combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
$|A|=6$
$ Pr(A) = \frac{6}{36} = \frac{1}{6}$
Out of 36 possibilities, a 3 on the first roll can be obtained 6 times: (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
$|B|=6$
$ Pr(B) = \frac{6}{36} = \frac{1}{6}$
$ Pr(A)=Pr(B)$