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Q.
Five different games are to be distributed among $4$ children randomly. The probability that each child get at least one game is
Probability
Solution:
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 $=\frac{n(E)}{n(S)}=\frac{240}{4^{5}}=\frac{15}{64}$