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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 ?

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  • $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$ Dec 15 '19 at 18:05
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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.

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There are three cases:

  1. A black ace and two other black cards that are not aces.
  2. A black ace, a black card that is not an ace, and a red card that is not an ace.
  3. 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 $$\color{red}{\binom{2}{1}}\binom{2}{1}\binom{24}{2} + \binom{2}{1}\binom{24}{1}\binom{24}{1} + \binom{2}{1}\binom{24}{2} = \color{red}{2808}$$

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