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. $\frac{55}{3}{\left(\frac{2}{3}\right)}^{11}$
• B. $55{\left(\frac{2}{3}\right)}^{11}$
• C. $22{\left(\frac{1}{3}\right)}^{11}$
• D. $228{\left(\frac{2}{3}\right)}^{12}$

Answer 1

The correct option is B $55{\left(\frac{2}{3}\right)}^{11}$

We are assuming that the given balls are different.
$n\left(S\right)=3^{12}$ since each ball will go in three ways.
We will assume that at least one of the boxes is having exactly 3 balls.
We can select $3$ balls from $12$ in ${}^{12}{C}_3=220$ ways.

We can select $1$ bag from $3$ in ${}^3{C}_1=3$ ways.
The rest $9$ balls can go in $2^9$ ways.

Hence, Probability $=\frac{2^9\times 220\times 3}{3^{12}}$

Probability $=55\times \frac{2^{11}}{3^{11}}$