Question

Five different games are to be distributed among 4 children randomly. The probability that each child get at least one game is

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Total number of ways of distribution is $4^5$.
$\therefore n\left(S\right)=4^5$
Total number of ways of distribution so that each child gets at least one game is
$4^5{-}^4{C}_1{3}^5{+}^4{C}_2{2}^5{-}^4{C}_3=1024-4\times 243+6\times 32-4=240\therefore n\left(E\right)=240$
Therefore, the required probability is
$\frac{n\left(E\right)}{n\left(S\right)}=\frac{240}{4^5}=\frac{15}{64}$