According to the question, a pair of dice has been thrown
So the number of elementary events in sample space will be $6^2=36$
n (S) = 36
By using the formula of probability, we get,
P (E) = favourable outcomes / total possible outcomes
(i) Let E be the event of getting odd number on first and 6 on second
E = {(1,6) (5,6) (3,6)}
n (E) = 3
P (E) = n (E) / n (S)
= 3 / 36
= 1/12
(ii) Let E be the event of getting greater than 4 on each die
E = {(5,5) (5,6) (6,5) (6,6)}
n (E) = 4
P (E) = n (E) / n (S)
= 4 / 36
= 1/9