# Twelve balls are distributed among three boxes. The probability that the first box contains three balls is

Question : Twelve balls are distributed among three boxes. The probability that the first box contains three balls is

(A) $\dfrac{110}{9}\left( \dfrac{2}{3}\right) ^{10}$

(B) $\dfrac{9}{100}\left( \dfrac{2}{3}\right) ^{10}$

(C) $\dfrac{12C_{3}}{12^{3}}\times 2^{9}$

(D) $\dfrac{12C_{3}}{3^{12}}$

the three boxes. Thus, there are $3^{12}$ ways in which 12 balls can be placed in 3 boxes. The number of ways in which 3 balls out of 12 can be put in the box is $12C_{3}$.
The remaining 9 balls can be placed in 2 boxes in $2^{9}$ ways, so that the required probability is,
$\dfrac{12C_{3}}{3^{12}}\times 2^{9}=\dfrac{110}{9}\left( \dfrac{2}{3}\right) ^{10}$