We are given that head comes on one side. So, there are only two possibilities- the other side is head ($\text{HH}$ coin) or the other side is tail ($\text{HT}$ coin). Favorable case is only $\text{HT}$, and hence answer is $\frac{1}{2}$.
Well, the above explanation is wrong because it assumes uniform probability distribution for the two cases $\text{HH}$ and $\text{HT}$. This is true initially as all 3 coins have equal probability of being chosen. But we are given that "head" comes in one face. $\text{HH}$ coin has two head faces, and hence has twice the chance of being chosen over $\text{HT}$ as we got a head face. i.e., for the two coins $\text{HH}$ and $\text{HT}$, the respective probabilities are $\frac{2}{3}$ and $\frac{1}{3}$ respectively given that one side is a head. Now, our required probability is simply $P(\{\text{HT}\}) = \frac{1}{3}$.
We can also apply Bayes' theorem but I guess solving intuitively is more fun :)