Question

The probability that when $12$ balls are distributed among three boxes, the first will contain three balls is

• A. $\frac{2^9}{3^{12}}$
• B. $\frac{{}^{12}{C}_2\times 2^9}{3^{12}}$
• C. $\frac{{}^{12}{C}_3\times 2^{12}}{3^{12}}$
• D. $\frac{{}^{12}{C}_3}{{12}^3}$

Answer 1

The correct option is B $\frac{{}^{12}{C}_2\times 2^9}{3^{12}}$

We have,

Boxes$=3$

Balls=$12$

Thus, There are $3$ ways in which a ball can be placed in any one of $3$ boxes.

Thus, there are $3^{12}$ ways in which $12$ balls can be placed in $3$ boxes.

Required probability$=\frac{{}^{12}{C}_3}{3^{12}}\times 2^9$.