TITLE: Probability of drawing exactly $1$ ace upon drawing $2$ cards from a deck
QUESTION [9 upvotes]: 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?

REPLY [3 votes]: 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 $384/2652 = 32/221$.