Zach has twelve identical-looking chocolate eggs. Exactly three of the eggs contain a special prize inside. Zach randomly gives three of the twelve eggs to each of Vince, Wendy, Xin, and Yolanda. What is the probability that only one child will receive an egg that contains a special prize (that is, that all three special prizes go to the same child)?
First, we note that $\binom{12}{3}=\frac{12!}{3!\cdot 9!}=\frac{12 \cdot 11 \cdot 10}{3 \cdot 2 \cdot 1}=2 \cdot 11 \cdot 10=220$.

This means that there are 220 different combinations of 3 eggs that can be chosen from 12 eggs. Zach randomly gives 3 eggs to each of the 4 children.

The probability that a specific one of these 4 children receives the specific set of all 3 special eggs is $\frac{1}{220}$, since there are 220 different combinations of 3 eggs that can be given and each set is equally likely to be chosen.

The probability that any one of these 4 children receives a specific set of 3 eggs is $4 \cdot \frac{1}{220}=\frac{1}{55}$ since the probability that each one receives the 3 special eggs is $\frac{1}{220}$ and these four events are disjoint.

Therefore, the probability that only one child receives an egg that contains a prize is $\frac{1}{55}$.