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 sum greater than 9
E = {(4,6) (5,5) (5,6) (6,4) (6,5) (6,6)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6
(ii) Let E be the event of getting even on first die
E = {(2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
n (E) = 18
P (E) = n (E) / n (S)
= 18 / 36
= 1/2