Q. $5$ different games are to be distributed among $4$ children randomly. The probability that each child gets atleast one game is
NTA AbhyasNTA Abhyas 2022
Solution:
Total cases $=4^{5}=1024$
Favourable cases $=\frac{5 !}{\left(1 !\right)^{3} 2 !}\times \frac{1}{3 !}\times 4!$ (By grouping)
$=240$
Hence, required probability
$=\frac{240}{1024}=\frac{15}{64}$
Favourable cases $=\frac{5 !}{\left(1 !\right)^{3} 2 !}\times \frac{1}{3 !}\times 4!$ (By grouping)
$=240$
Hence, required probability
$=\frac{240}{1024}=\frac{15}{64}$