Probability of drawing exactly $1$ ace upon drawing $2$ cards from a deck

We have: $$P_{\text{ only one ace }} = P_{\text{ ace }1} P_{\text{ non-ace }2} + P_{\text{ non-ace }1} P_{\text{ ace }2} = \frac{4}{52}\times \frac{48}{51} + \frac{48}{52}\times \frac{4}{51} = \frac{32}{221}$$

You forgot to assume $P(B)\times P(A|B)=\frac{48}{52}\times\frac4{51}=\frac{16}{221}$.

So $P(A)\times P(B|A)+P(B)\times P(A|B)= \frac{16}{221}+ \frac{16}{221}= \frac{32}{221}$.

The correct answer is $4$ , i.e. $\frac{32}{221}$

Answer 4 seems correct to me because:

$$P=\frac {\binom4 1\binom {48} 1}{\binom {52} 2}=\frac {32}{221}$$