Probability of drawing exactly $1$ ace upon drawing $2$ cards from a deck I got this question and answered it incorrectly. I haven't yet seen the correct answer. The possible answers were:

  
*
  
*$\frac{4}{52}$
  
*$\frac{16}{221}$ (my answer)
  
*$\frac{2}{52}$
  
*$\frac{32}{221}$
  


My reasoning is the following:
Event A: Card is not an ace.
Event B: Card is an ace.
$$P(A)\times P(B|A)=\frac{4}{52}\times \frac{48}{51}=\frac{16}{221}$$
This is assuming the cards are drawn sequentially. Drawing exactly one ace from a single draw is $4/52$ but to ensure that only a single ace was drawn one should consider the probability of not getting a second ace.
How am I wrong?
 A: Answer 4 seems correct to me because: 
$$P=\frac {\binom4 1\binom {48} 1}{\binom {52} 2}=\frac {32}{221}$$
A: You can figure this out without any knowledge of conditional probabilities:
There are 52 choices for the first pick and 51 choice for the second pick. So, in total, there are $52 \times 51 = 2652$ possible "hands" that consist of two cards.
Now, the question is, how many of these 2652 contain a single ace. 
Let's do some counting of the possible one-ace hands:
There are $4 \times 48 = 192$ hands that have an ace as their first pick and a non-ace as the second pick (because there are 4 aces and 48 non-aces).
Similarly, there are $48 \times 4 = 192$ hands that have a non-ace as their first pick and an ace as the second pick. So the total number of hands that have a single ace is $192+192=384$.
So, the probability of a one-ace hand is $384/2652$. Dividing top and bottom by $12$ gives us $384/2652 = 32/221$.
A: We have: $$P_{\text{ only one ace }} = P_{\text{ ace }1} P_{\text{ non-ace }2} + P_{\text{ non-ace }1} P_{\text{ ace }2} = \frac{4}{52}\times \frac{48}{51} + \frac{48}{52}\times \frac{4}{51} = \frac{32}{221}$$
A: 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. $\frac{32}{221}$
