Question

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

• A. $\frac{2^9}{3^{12}}$
• B. $\frac{{{}^{12}{C}_{3}2}^9}{3^{12}}$
• C. $\frac{{{}^{12}{C}_{3}2}^{12}}{3^{12}}$
• D. None of these

Answer 1

Answer: (B)

$\frac{{{}^{12}{C}_{3}2}^9}{3^{12}}$

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