Let 9 distinct balls be distributed among 4 b | vedclass

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(C) The total number of ways to distribute $9$ distinct balls into $4$ boxes is $4^{9}$.The number of ways to choose $3$ balls for box $B_{3}$ is ${}^

Vedclass · Accepted Answer

$\{x \in R : |x-3| < 1\}$

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(C) The total number of ways to distribute $9$ distinct balls into $4$ boxes is $4^{9}$.The number of ways to choose $3$ balls for box $B_{3}$ is ${}^{9}C_{3}$.The remaining $6$ balls can be distributed among the other $3$ boxes $(B_{1}, B_{2}, B_{4})$ in $3^{6}$ ways.Thus,the probability that $B_{3}$ contains exactly $3$ balls is $P = \frac{{}^{9}C_{3} \cdot 3^{6}}{4^{9}}$.We can rewrite this as $P = \frac{{}^{9}C_{3} \cdot 3^{6}}{4^{9}} = \frac{84 \cdot 3^{6}}{4^{9}}$.Since we want the form $k \left(\frac{3}{4}\right)^{9}$,we write $P = k \cdot \frac{3^{9}}{4^{9}}$.Equating the two expressions: $k \cdot \frac{3^{9}}{4^{9}} = \frac{84 \cdot 3^{6}}{4^{9}}$.$k = \frac{84 \cdot 3^{6}}{3^{9}} = \frac{84}{3^{3}} = \frac{84}{27} = \frac{28}{9} \approx 3.11$.Checking the options for $k = \frac{28}{9} \approx 3.11$:$|x-3| < 1 \Rightarrow 2 < x < 4$. Since $3.11$ lies in this interval,option $C$ is correct.

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:

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