probability at least two of the cards to be black (spades or clubs) and exactly one card to be an ace So we draw 3 cards.
I am looking for the probability at least two of the cards to be black (spades or clubs) and exactly one card to be an ace.
So my attempt is :
1. case: The ace is black. I choose the ace, then the other black card, then the other card (shouldn't be an other ace) :
$$\binom{2}{1}\binom{24}{1}\binom{47}{1}$$
2. case The ace is not black. I choose the ace, then the other 2 black cards (they cannot be aces)
$$\binom{2}{1}\binom{24}{2}$$
So I get $$\frac{2808}{\dbinom{52}{3}}$$  which is NOT the answer. Any tips what I am doing wrong ?
 A: There are three cases:


*

*A black ace and two other black cards that are not aces.

*A black ace, a black card that is not an ace, and a red card that is not an ace.

*A red ace and two black cards that are not aces.


A black ace and two other black cards are not aces: Choose one of the two black aces and two of the $24$ black cards that are not aces.
$$\binom{2}{1}\binom{24}{2}$$
A black ace, a black card that is not an ace, and a red card that is not an ace: Choose one of the two black aces, one of the $24$ black cards that are not aces, and one of the $24$ red cards that are not aces.
$$\binom{2}{1}\binom{24}{1}\binom{24}{1}$$
A red ace and two black cards that are not aces:  Choose one of the two red aces and two of the $24$ black cards that are not aces.
$$\binom{2}{1}\binom{24}{2}$$
Thus, the number of favorable cases is
$$\binom{2}{1}\binom{24}{2} + \binom{2}{1}\binom{24}{1}\binom{24}{1} + \binom{2}{1}\binom{24}{2}$$
so the probability of drawing at least two black cards and exactly one ace when three cards are drawn randomly from a standard deck is 
$$\frac{\dbinom{2}{1}\dbinom{24}{2} + \dbinom{2}{1}\dbinom{24}{1}\dbinom{24}{1} + \dbinom{2}{1}\dbinom{24}{2}}{\dbinom{52}{3}}$$
You counted each case in which a black ace and two black non-aces were drawn twice, once for each way of designating one of the black non-aces as the black non-ace.  Notice that 
$$\color{red}{\binom{2}{1}}\binom{2}{1}\binom{24}{2} + \binom{2}{1}\binom{24}{1}\binom{24}{1} + \binom{2}{1}\binom{24}{2} = \color{red}{2808}$$
A: I think you should consider three cases and add them together. 
(2/52* 24/51* 26/50) + ( 24/52* 2/51* 26/50) + ( 24/52* 23/51* 2/50). 
Also I would consider the three different combinations. 
