An urn contains 25 balls of which I0 balls bear a mark 'X' and 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. atlest one ball will bear 'Y' mark
iv. the number of balls with 'X' marks and 'Y' mark will be equal

Answer

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