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