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If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls, is

$\frac{55}{3}\left(\frac{2}{3}\right)^{11}$

$\frac{55}{3}\left(\frac{2}{3}\right)^{10}$

$220\left(\frac{1}{3}\right)^{12}$

$22\left(\frac{1}{3}\right)^{11}$

Option A

There seems to be ambiguity in the question. It should be mentioned that boxes are different and one particular box has 3 balls

Then , number of ways= $\frac{^{12}C_{3}\times 2^{9}}{3^{12}}$

= $\frac{56}{3}\left(\frac{2}{3}\right)^{11}$

According to the question,

$\frac{^{3}C_{1}\times ^{12}C_{3}2^{9}-^{3}C_{2}^{12}C_{3}^{9}C_{3}+\frac{12!\times3!}{3!3!6!3!}}{3^{12}}$

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