Total numbers of elementary events are: 6 x 6 = 36

(i) Let E be the event of getting number other than 5 on both dices

The Cases where 5 comes up on at least one time are (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6) and (6, 5).

The number of cases = 11 being combinations with 5 on at least one dice

The number of favourable cases where 5 will not come up either time = 36 – 11 = 25.

∴ P (5 will not come up either time) = P (E) = 25/36

(ii) Let E be the event of getting one number as 5 on either of dice

The favourale outcomes where 5 comes up on exactly one time are: (1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6) and (6, 5).

The number of favourable such cases = 10.

∴ P (5 will come up exactly one time) = P (E) = 10/36 =5/18

(iii) Favourable event when 5 come up on exactly two times is (5, 5).

The number of such cases = 1.

∴ P (5 will come up both the times) = P (E) = 1/36