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}

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