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  • An urn contains 25 balls of which 10 balls bear a mark 'X' and the remaining 15 bear a mark 'Y' (iv) the number of balls with 'X' mark and 'Y' mark will be equal.

Q. 5   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 

            (iv) the number of balls with 'X' mark and 'Y' mark will be equal.

Answers (1)

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P(equal\, to\, the \, number\, of\, balls \, with \, X \, mark \, and\, Y \, mark )=P(Z=3)

                                      P(Z=3)=^6C_3.(\frac{2}{5})^2.(\frac{3}{5})^3

                                    P(Z=3)=\frac{20\times 8\times 27}{15625}

                                             P(Z=3)=\frac{864}{3125}

Posted by

seema garhwal

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