Refer to Exercise 1 above.

According to the solution of exercise 1, we have

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}and n(6) and n(S) = 6 x 6 = 36

So, P(A) = n(A)/n(S) = 6/36 = 1/6

And, B = {(4, 6), (6, 4), (5, 5), (5, 6), (6, 5), (6, 6)}; n(B) = 6 and n(S) = 36

So, P(B) = n(B)/n(S) = 6/36 = 1/6

Now, A ⋂ B = {(5, 5), (6, 6)}

So, P(A ⋂ B) = 2/36 = 1/18

Hence, if A and B are not independent, then

P(A ⋂ B) ≠ P(A).P(B)

1/18 ≠ 1/6 x 1/6 ⇒ 1/18 ≠ 1/36

Therefore, A and B are not independent events.