Answer:
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:-)
