A standard fair dice is rolled twice. What is the probability of getting an odd number on the first roll and any number except 4 on the other?

Let, O and P be the events of getting odd number and any other number except 4 respectively.

n (S) = 6

O = (1,3,5) so, n (O) = 3

P = (1,2,3,5,6) so, n (P) = 5

So, probability of getting an odd number and any other number except 4 will be

n (O) / n (S) * n (P) / n (S)

=3/6 * 5/6

=5/12

May be it's the answer I guess. Hope you're clear with it:-)