Question

# If 12 distinct balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is (a) $\dfrac{55}{3}{{\left( \dfrac{2}{3} \right)}^{11}}$ (b) $55{{\left( \dfrac{2}{3} \right)}^{11}}$(c) $22{{\left( \dfrac{1}{3} \right)}^{11}}$(d) $228{{\left( \dfrac{2}{3} \right)}^{11}}$

Hint: To solve this question, we will first find the number of ways in which the given 12 balls can go in any one of the three boxes. This will be our sample set. Then we will find the number of ways in which we can choose 3 balls out of 12. We will also find the number of ways in which we can choose one of the three boxes. Then the rest of the remaining balls can go in any of the two remaining boxes. We will know the number of ways in which the remaining balls can be placed. The product of the number of ways to choose 3 balls out of 12 and number of ways of choosing one box out of 3 and number of ways remaining balls can be placed will be equal to the number of ways exactly 3 balls will be placed in one of the boxes. Thus, we can find the probability as the ratio of number of ways to place exactly 3 balls in one of the boxes and number of ways placing 12 balls in 3 boxes randomly.

Let S be the event of placing 12 distinct balls in 3 boxes randomly.
Each ball can be placed in 3 ways and there are 12 such balls.
Thus, number of ways 12 balls can be placed in 3 boxes is n(S) = ${{3}^{12}}......\left( 1 \right)$
Now, let A be the event of placing 3 balls in exactly one of the boxes.
To find the number of elements in the event set A, first we have to choose any 3 balls from the available 12 balls.
This can be done in $^{12}{{C}_{3}}$ = 220 ways.
Then, we have to choose one of the available 3 boxes. This is done in $^{3}{{C}_{1}}=3$ ways.
The remaining 9 balls can go into 1 of the 2 boxes. Hence the number of ways 9 balls can be placed in 2 boxes is given as ${{2}^{9}}$.
Therefore, number of elements in event set A is n(A) = ${{2}^{9}}\times 3\times 220......\left( 2 \right)$
Thus, the probability of event A happening, that one of the boxes has exactly 3 boxes is given as $\Rightarrow P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$
We shall find the values of n(A) and n(S) from (1) and (2).
\begin{align} & \Rightarrow P\left( A \right)=\dfrac{{{2}^{9}}\times 3\times 220}{{{3}^{12}}} \\ & \Rightarrow P\left( A \right)=\dfrac{{{2}^{9}}\times 4\times 55}{{{3}^{11}}} \\ & \Rightarrow P\left( A \right)=55\left( \dfrac{{{2}^{11}}}{{{3}^{11}}} \right) \\ \end{align}

So, the correct answer is “Option b”.

Note: If we have to choose r items from n available items, then it can be in $^{n}{{C}_{r}}$ ways, where $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.