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`

Options

A. `1//4`
B. `15//64`
C. `21//64`
D. none of these

Answer

B

Explanation

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)`