An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y'. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear 'X' mark.
(ii) not more than 2 will bear 'Y' mark.
(iii) at least one ball will bear 'Y' mark.
 (iv) the number of balls with 'X' mark and 'Y' mark will be equal.

Answer

(i) `((2)/(5))^(6)` (ii) `7((2)/(5))^(4)` (iii) `1-((2)/(5))^(6)` (iv) `(864)/(3125)`