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(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) $228{{\left( \dfrac{2}{3} \right)}^{11}}$

Answer
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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}$