Question

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

• A. 1 / 4
• B. 15 / 64
• C. 21 / 64
• D. None of these

Answer 1

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
$=4^5-{}^4{C}_1{3}^5+{}^4{C}_2{2}^5-{}^4{C}_3$
$=1024-4\times 243+6\times 32-4=240$
$\therefore n(E)=240$
Therefore, the required probability $=\frac{n(E)}{n(S)}=\frac{240}{4^5}=\frac{15}{64}$