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}}
Answer : option (A)
Solution :
Since each ball can be placed in any one of the 3 boxes, therefore there are 3 ways in which a ball
can be placed in any one therefore there are 3 ways in which a ball can be placed in any one of
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}