MCQ
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 $\mathrm{k}$ lies in the set:
  • A
    $\{x \in R:|x-5| \leq 1\}$
  • B
    $\{x \in R:|x-2| \leq 1\}$
  • $\{x \in R:|x-3|<1\}$
  • D
    $\{x \in R:|x-1|<1\}$

Answer

Correct option: C.
$\{x \in R:|x-3|<1\}$
c
$\text { required probability }=\frac{{ }^{9} \mathrm{C}_{3} \cdot 3^{6}}{4^{9}}$

$=\frac{{ }^{9} \mathrm{C}_{3}}{27} \cdot\left(\frac{3}{4}\right)^{9}$

$=\frac{28}{9} \cdot\left(\frac{3}{4}\right)^{9} \Rightarrow \mathrm{k}=\frac{28}{9}$

Which satisfies $|x-3|\,<\,1$

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