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 even on one and multiple of three on other
E = {(2,3) (2,6) (4,3) (4,6) (6,3) (6,6) (3,2) (3,4) (3,6) (6,2) (6,4)}
n (E) = 11
P (E) = n (E) / n (S)
= 11 / 36
(ii) Let E be the event of getting neither 9 or 11 as the sum
E = {(3,6) (4,5) (5,4) (5,6) (6,3) (6,5)}
n (E) = 6
P (E) = n (E) / n (S)
= 6 / 36
= 1/6