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) `2//5` b. `12//25` c. `3//5` d. none of these

Options

A. `2//5`
B. `12//25`
C. `3//5`
D. none of these

Answer

C

Explanation

We have,
`n(S) = 5^(5)`
For computing favorable 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![(5!)/(2! 2! 2!) + (5!)/(3! 2!)] = 150` ways
`implies n(A) = .^(5)C_(2) xx 150`
Hence,
`P(E) = .^(5)C_(2) xx (150)/(5^(5)) = (60)/(125) = (125)/(25)`