Question

$12$ balls are distributed among $3$ boxes. The probability that the 1st box contains $3$ balls is

• A. $(\frac{2}{3}{)}^{10}$
• B. $\frac{100}{9\times 3^{10}}$
• C. $\frac{110}{9}\times (\frac{2}{3}{)}^{10}$
• D. $\frac{100}{9}$

Answer 1

Answer: (C)

$\frac{110}{9}\times (\frac{2}{3}{)}^{10}$

The each ball has chance of going into any one of the three boxes
Therefore total ways of distributing $12$ balls in $3$ boxes $=(3{)}^{12}$
No. of ways of choosing three balls out of $10$ to put in first box ${=}^{12}{C}_3$
ways of distributing $9$ balls in $2$ boxes $=(2{)}^9$
Probability $={}^{12}{C}_3\ast \left(2{)}^9/\right(3{)}^{1}2=110/9\ast (2/3{)}^{10}$