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 a doublet of prime numbers

E = {((2, 2) (3, 3) (5, 5)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12

(ii) Let E be the event of getting a doublet of odd numbers

E = {(1, 1) (3, 3) (5, 5)}

n (E) = 3

P (E) = n (E) / n (S)

= 3 / 36

= 1/12