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An urn contains 25 balls of which 10 bea...

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

Text Solution

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1) `P=(10/25)(10/25)(10/25)....(1-/25)=(2/5)^6`
2)P(not more than 2)=`P(0Y)+P(1Y)+P(2Y)`
`=(2/5)^6+6C_1*3/5(2/5)^5+6c_2(3/5)^2(2/5)^4`
`=(2/5)^4[1+6*3/5*2/5+15*9/5]`
`=(2/5)^4(136/25)=34(2/5)^6`.
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