In a single throw with 2 dices, what is probability of neither getting an even number on one and nor a multiple of 3 on other?

We first calculate the probability of getting an even number on one and a multiple of 3 on other,Here, n(s) = 6x6 = 36 and

E = (2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4)(3,6) (6,2)(6,4)

n(E) = 11P(E) = 11/36Required probability = 1-11/36 = 25/36