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)`