InterviewSolution
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InterviewSolution
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Five different marbles are placed in 5 different boxes randomly. Thenthe probability that exactly two boxes remain empty is (each box can hold anynumber of marbles)`2//5`b. `12//25`c. `3//5`d. none of theseA. `2//5`B. `12//25`C. `3//5`D. none of these |
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Answer» Correct Answer - C 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)` |
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