Question

Twelve balls are distributed among three boxes, find the probability that the first box will contain three balls.
$A.$ $\dfrac{{110}}{9}{\left( {\dfrac{2}{3}} \right)^{10}}$
$B.$ $\dfrac{9}{{110}}{\left( {\dfrac{2}{3}} \right)^{10}}$
$C.$ $\dfrac{{{}^{12}{C_3}}}{{{{12}^3}}} \times {2^9}$
$D.$ $\dfrac{{{}^{12}{C_3}}}{{{3^{12}}}}$

Answer 1

Hint – In this particular question, firstly we find the total number of probable cases then we find out the number of favourable cases. Then according to the given conditions we can simply find the required probability.

Complete step-by-step answer:
Since each ball can be placed in any one of the 3 boxes.
Thus, there are 3 ways in which a ball can be placed in any one of 3 boxes.
Therefore, the total number of ways in which 12 boxes can be put in to three boxes $ = {3^{12}}$
Again, out of 12 given balls, 3 balls can be chosen in ${}^{12}{C_3}$ ways.
Consider that the above chosen 3 balls are put into the first box. So, the remaining 12-3$ = 9$ balls are to be out in the other two boxes.
It can be done ${2^9}$ ways.
Therefore, the total number of ways in which 3 balls are put in the first box and remaining 9 in other two boxes $ = {}^{12}{C_3} \times {2^9}$ ways
Hence, the probability that first box will contain 3 balls $ = \dfrac{{{}^{12}{C_3} \times {2^9}}}{{{3^{12}}}}$
$ = \dfrac{{110}}{9}{\left( {\dfrac{2}{3}} \right)^{10}}$

Note – In order to solve such types of questions, one must know that by dividing the number of successes by the total number of attempts. The result will be a number between 0 and 1, and they can also be expressed as percentages ranging from 0% to 100%.