If 4 different biscuts are distributed among 3 children at random, the probability that the first child receives exactly one biscut is

4 different biscuts are distributed among 3 children at random

$n\left(S\right)=3^4=81$

E= Event in which the first child receives exactly one biscut 

$n\left(E\right){=}^4{C}_1{\left(2\right)}^3=32$

$P\left(E\right)=\frac{32}{81}$