InterviewSolution
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InterviewSolution
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Five different games are to be distributed among 4 children randomly.The probability that each child get at least one game is`1//4`b. `15//64`c. `5//9`d. `7//12`A. `1//4`B. `15//64`C. `21//64`D. none of these |
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Answer» Correct Answer - B Total number of ways of distribution is `4^(5)`. `therefore n(S) = 4^(5)` Total number of ways of distribution so that each child gets at least one game is `4^(5) - .^(4)C_(1) 3^(5) + .^(4)C_(2) 2^(5) - .^(4)C_(3) = 1024 - 4 xx 243 + 6 xx 32 -4 = 240` `therefore n(E) = 240` Therefore, the required probability is `(n(E))/(n(S)) = (240)/(4^(5)) = (15)/(64)` |
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