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Question

(A) Each of the $12$ balls can be placed in any of the $3$ boxes in $3$ ways. Therefore,the total number of ways to distribute $12$ balls into $3$ box

Vedclass · Accepted Answer

$\frac{{}^{12}C_3 	imes 2^9}{3^{12}}$

Vedclass · Answer

(A) Each of the $12$ balls can be placed in any of the $3$ boxes in $3$ ways. Therefore,the total number of ways to distribute $12$ balls into $3$ boxes is $3^{12}$.To find the number of favorable outcomes where the first box contains exactly $3$ balls:$1$. Choose $3$ balls out of $12$ to be placed in the first box,which can be done in ${}^{12}C_3$ ways.$2$. The remaining $9$ balls must be placed in the other $2$ boxes. Each of these $9$ balls has $2$ choices,so there are $2^9$ ways to distribute them.Thus,the number of favorable ways is ${}^{12}C_3 	imes 2^9$.The required probability is $\frac{	ext{Favorable outcomes}}{	ext{Total outcomes}} = \frac{{}^{12}C_3 	imes 2^9}{3^{12}}$.

$12$ balls are distributed among $3$ boxes. The probability that the first box will contain exactly $3$ balls is:

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