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You are to pick 4 cards randomly from a condensed deck of cards that contains four suits - $\color{red}{\heartsuit,\diamondsuit},\clubsuit,\spadesuit$ - and the following denominations: Ace, $2, 3, 4, 5, 6, 7, 8, 9$, and $10$. There are no face-cards in this deck.

What is the probability you get two aces and two hearts?

I'm really not sure about this question. At first I assumed that the aces could be one of the hearts, so there's $3$ ways to pick two aces that aren't hearts. Then, there are nine cards left that we can viably pick that aren't aces, that we could pick hearts from ${9 \choose 2} = 36$, so $36 \cdot 3 = \frac{108}{91390}$, but apparently that isn't the right answer.

How would this calculation differ if I was to include the aces?

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The denominator is the number of ways we can select $4$ of the $40$ cards in the deck.

There are two cases to consider:

  1. Neither ace is a heart: You handled this case correctly.
  2. One ace is a heart: Observe that there are three ways to select the other ace and that there are nine ways to select the other heart. Assuming the question means exactly two aces and two hearts, the fourth card in the hand must be selected from one of the $40 - 4 - 9 = 27$ cards that is neither an ace nor a heart.

The numerator is found by adding the above cases.

$$\frac{\dbinom{3}{2}\dbinom{9}{2} + \dbinom{3}{1}\dbinom{9}{1}\dbinom{27}{1}}{\dbinom{40}{4}}$$

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