If 12 different balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is :
Solutions · 1
Official solution
1st ball can go any of the 3 boxes. So total choices for 1st ball = 3
2nd ball can also go any of the 3 boxes. So total choices for 2nd ball = 3
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12th ball can go any of the 3 boxes. So total choices for 12th ball = 3
Total choices for all 12 balls = \[3 \times \] 3×\[3 \times \].................12 times = 312.
Now question says choose 3 balls from 12 balls. So no of ways = 12C3 ways.
And then put it in a box. No of ways we can put = \[{}^{12}{C_3} \times 1\] ways.
Now we have 9 balls left and we have to put those 9 balls in the remaining 2 boxes.
Each ball can go to any of the 2 boxes, so for each ball there is 2 choices.
\[\therefore\] Total ways for 9 balls = 29
\[\therefore\] Total ways we can put those 12 balls in the boxes = \[{}^{12}{C_3} \times 1 \times {2^9}\]
\[\therefore\] Required probability = \[{{{}^{12}{C_3} \times 1 \times {2^9}} \over {{3^{12}}}}\] = \[{{55} \over 3}{\left( {{2 \over 3}} \right)^{11}}\]
So option (C) is correct.