Question

Five different marbles are placed in 5 different boxes randomly. Then the probability that exactly two boxes remain empty is (each box can hold any number of marbles).

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

We have, the number elements in sample space$=n\left(S\right)=5^5$ as placing 5 different marbles in 5 different boxes.
For computing favourable outcomes, 2 boxes which are to remain empty, can be selected in ${}^5{C}_2$ ways and 5 marbles can be placed in the remaining 3 boxes in groups of 221 or 311 in
$3![\frac{5!}{2!2!2!}+\frac{5!}{3!2!}]$
$=(3\times 2\times 2)[\frac{5\times 4\times 3\times 2!}{2\times 2\times 2}+\frac{5\times 4\times 3!}{3!\times 2}]$
$=6\times [15+10]$
$=150ways$

Thus, number of elements in favourable event, $n\left(A\right){=}^5{C}_2\times 150$
Hence,$P\left(E\right)=\frac{n\left(A\right)}{n\left(S\right)}{=}^5{C}_2\times \frac{150}{5^5}=\frac{60}{125}=\frac{12}{25}$