Question

If $10$ different balls has to placed in $4$ distinct boxes at random, then the probability that two of these boxes contain exactly $2$ and $3$ balls is :

• A. $\frac{965}{2^{10}}$
• B. $\frac{945}{2^{10}}$
• C. $\frac{945}{2^{11}}$
• D. $\frac{965}{2^{11}}$

Answer 1

The correct option is B $\frac{945}{2^{10}}$

Total ways to distribute $10$ balls in $4$ boxes is $=4^{10}$
Total ways of placing exactly $2$ and $3$ balls in any two of these boxes is $={}^4{C}_2\times {}^{10}{C}_5\times \frac{5!}{2!\times 3!}\times 2\times 2^5$
$P\left(E\right)=\frac{{}^4{C}_2\times {}^{10}{C}_5\times \frac{5!}{2!\times 3!}\times 2\times 2^5}{2^{20}}=\frac{945}{2^{10}}$