# 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)

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

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,

$P\left( E\right) =5C_{2}\times \dfrac{150}{5^{5}}=\dfrac{60}{125}=\dfrac{12}{25}$
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