Question

Tweleve balls are distributed among three boxes. The probability that the first box contains three balls is

• A. $\frac{110}{9}(\frac{2}{3}{)}^{10}$
• B. $\frac{9}{110}(\frac{2}{3}{)}^{10}$
• C. $\frac{{}^{12}{C}_3}{{12}^3}\times 2^9$
• D. $\frac{{}^{12}{C}_3}{3^{12}}$

Answer 1

The correct option is A $\frac{110}{9}(\frac{2}{3}{)}^{10}$

Since each ball can be placed in any one of the three boxes, therefore there are $3$ ways in which a ball can be placed in any one of the three boxes. Thus, there are $3^{12}$ ways in which $12$ can be put in these 3 boxes but the condition is that first box should contain 3 balls. So, 3 balls can be selected from 12 balls in ${}^{12}{C}_3$ ways. The remaining $9$ can be placed in $2$ boxes in $2^9$ ways.
So, required probability is
$\frac{{}^{12}{C}_3}{3^{12}}\times 2^9=\frac{110}{9}(\frac{2}{3}{)}^{10}$