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 

9 balls are be distributed among 4 boxes ${\mathrm{B}}_1, {\mathrm{B}}_2, {\mathrm{B}}_3, {\mathrm{B}}_4$

$\mathrm{n}\left(\mathrm{S}\right)=4^9$

E=box ${\mathrm{B}}_3$ Contain exactly 3 balls

$\begin{array}{rcl}\mathrm{n}\left(\mathrm{E}\right)& =& {}^9{\mathrm{C}}_3\times 36\\ \mathrm{P}\left(\mathrm{E}\right)& =& \frac{{}^9{\mathrm{C}}_3\times 3^6}{4^9}=\frac{9\times 8\times 7}{3\times 3\times 3\times 6} {\left(\frac{3}{4}\right)}^9\\ \mathrm{k}& =& \frac{56}{3\times 6}=\frac{25}{9}=3.3\end{array}$

$\begin{array}{c}|\mathrm{x}-1|<1\\ \Rightarrow 0<\mathrm{x}<2\end{array} \overline{) \begin{array}{c}|\mathrm{x}-5|\le 1\\ 4\le \mathrm{x}\le 6\end{array} \overline{) \begin{array}{c}|\mathrm{x}-3|<1\\ 2<\mathrm{x}<4\\ \mathrm{correct}\end{array}}}$