Question : Five different marbles are placed in 5 different boxes randomly. Then the probability that exactly two boxes remain empty is (each box can any number of marbles)
(A) 2/5
(B) 12/25
(C) 3/5
(D) none of these
Answer : option (B)
Solution :
we have,
n\left( S\right) =5^{5}For computing favourable outcomes, 2 boxes which are to remain empty, can be selected in 5C_{2} ways and 5 marbles can be placed in the remaining 3 boxes in groups of 221 or 311 in
\begin{aligned}3!\left[ \dfrac{5!}{2!2!2!}+\dfrac{5!}{3!2!}\right] =150ways\ \Rightarrow n\left( A\right) =5C_{2}\times 150\end{aligned}Hence,
P\left( E\right) =5C_{2}\times \dfrac{150}{5^{5}}=\dfrac{60}{125}=\dfrac{12}{25}