Question

Let $9$ distinct balls be distributed among $4$ boxes, $B_1,B_2,B_3$ and $B_4$. If the probability that $B_3$ contains exactly $3$ balls is $k{\left(\frac{3}{4}\right)}^9$, then $k$ lies in the set

• A. $\{x\in R:|x-3|<1\}$
• B. $\{x\in R:|x-1|<1\}$
• C. $\{x\in R:|x-5|\le 1\}$
• D. $\{x\in R:|x-2|\le 1\}$

Answer 1

The correct option is A $\{x\in R:|x-3|<1\}$

Number of ways to distribute $9$ distinct balls in $4$ boxes $=n\left(S\right)=4^9$
Number of ways of favourable distribution
$=n\left(E\right)={}^9{C}_3\times 3^6=\frac{10\times 8\times 7}{3\times 2\times 1}\times 3^6=28\cdot 3^7$

$\therefore$ Required probability$=\frac{28\times 3^7}{4^9}=\frac{28}{9}\cdot {\left(\frac{3}{4}\right)}^9$
$\therefore k=\frac{28}{9}$
$\therefore |x-3|<1\Rightarrow x\in (2,4)$