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:-)

answer: x= -2y

step-by-step explanation:

the answer is 0.3175 hope that you ^w^

also you might actually want to put this picture up if you want people to solve it luckily i remember this question