Question

If $10$ different balls has to be 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

$\frac{945}{2^{10}}$

Explanation for correct option:

Step-1: Express given data.

Given, Total $10$ different ball. and $4$ different box.

Now, we have to find the probability that two of these boxes contain exactly $2$ and $3$ balls.

Step-2: Finding required probability:

Total number of ways to distribute $10$balls in $4$ different boxes ${\left(4\right)}^{10}={\left(2\right)}^{20}$.

Total number of ways that which two boxes contain exactly $2$ and $3$ balls is ${}^{4}C_2\times {}^{2}C_1\times {}^{10}C_2\times {}^{8}C_3\times {\left(2\right)}^5$

Now required probability is $\frac{\left({}^{4}C_2\times {}^{2}C_1\times {}^{10}C_2\times {}^{8}C_3\times {\left(2\right)}^5\right)}{{\left(2\right)}^{20}}$.

$=\frac{\left({\displaystyle \frac{4\times 3\times 2!}{\left(4-2\right)!\times 2!}}\times {\displaystyle \frac{2!}{1!\times 1!}}\times {\displaystyle \frac{10!}{\left(10-8\right)!\times 2!}}\times {\displaystyle \frac{8!}{\left(8-3\right)!\times 3!}}\times {\left(2\right)}^5\right)}{{\left(2\right)}^{20}}$

$=\frac{\left({\displaystyle \frac{4\times 3}{1\times 2}}\times {\displaystyle \frac{1\times 2}{1\times 1}}\times {\displaystyle \frac{10\times 9}{2}}\times {\displaystyle \frac{8\times 7\times 6}{3\times 2\times 1}}\times {\left(2\right)}^5\right)}{{\left(2\right)}^{20}}$

$=\frac{945}{{\left(2\right)}^{10}}$

Hence, correct answer is option $\left(B\right)$