Question:
Let 9 distinct balls be distributed among 4 boxes, $B_{1}, B_{2}, B_{3}$ and $B_{4}$. If the probability than $B_{3}$
contains exactly 3 balls is $\mathrm{k}\left(\frac{3}{4}\right)^{9}$ then $\mathrm{k}$ lies in the
set:
Correct Option: , 2
Solution:
required probability $=\frac{{ }^{9} \mathrm{C}_{3} \cdot 3^{6}}{4^{9}}$
$=\frac{{ }^{9} \mathrm{C}_{3}}{27} \cdot\left(\frac{3}{4}\right)^{9}$
$=\frac{28}{9} \cdot\left(\frac{3}{4}\right)^{9} \Rightarrow \mathrm{k}=\frac{28}{9}$
Which satisfies $|x-3|<1$
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