Question
If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes contains exactly 3 balls is:
A.
$$220{\left( {\frac{1}{3}} \right)^{12}}$$
B.
$$22{\left( {\frac{1}{3}} \right)^{11}}$$
C.
$$\frac{{55}}{3}{\left( {\frac{2}{3}} \right)^{11}}$$
D.
$$55{\left( {\frac{2}{3}} \right)^{10}}$$
Answer :
$$\frac{{55}}{3}{\left( {\frac{2}{3}} \right)^{11}}$$
Solution :
Note :- The question should state ‘3 different’ boxes instead of ‘3 identical boxes’ and one particular box has 3 balls. Then the solution would be:
Required probability $$ = \frac{{^{12}{C_3} \times {2^9}}}{{{3^{12}}}} = \frac{{55}}{3}{\left( {\frac{2}{3}} \right)^{11}}$$