Question

$5$ different games are to be distributed among $4$ children randomly. The probability that each child gets atleast one game is

• A. $\frac{1}{4}$
• B. $\frac{15}{64}$
• C. $\frac{21}{64}$
• D. $\frac{17}{632}$

Answer 1

Answer: (B)

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
$=\frac{240}{1024}=\frac{15}{64}$