Question

Twelve balls are distributed among three boxes. The probability that the first box will contains three balls.

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

Answer 1

The correct option is B $\frac{{}^{12}{C}_3\times 2^9}{3^{12}}$

Since each ball can be placed into any one of the three boxes. So, the total number of ways in which $12$ balls can be placed into three boxes is $3^{12}$.
Out of $12$ balls, $3$ balls can be chosen in ${}^{12}{C}_3$ ways for first box.
Now, remaining $9$ balls can be placed in the remaining $2$ boxes in $2^9$ ways.
So, the total number of ways in which $3$ balls can be placed in the first box and the remaining balls in other two boxes is ${}^{12}{C}_3\times 2^9$.
Hence, required probability is
$\frac{{}^{12}{C}_3\times 2^9}{3^{12}}$