# Q. 5   An urn contains 25 balls of which 10 balls bear a mark $\inline 'X'$ and the remaining 15 bear a mark $\inline 'Y'.$ 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 $\inline 'X'$ mark and $\inline 'Y'$ mark will be equal.

$P(equal\, to\, the \, number\, of\, balls \, with \, X \, mark \, and\, Y \, mark )=P(Z=3)$
$P(Z=3)=^6C_3.(\frac{2}{5})^2.(\frac{3}{5})^3$
$P(Z=3)=\frac{20\times 8\times 27}{15625}$
$P(Z=3)=\frac{864}{3125}$