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?
Best Answer
The denominator is the number of ways we can select $4$ of the $40$ cards in the deck.
There are two cases to consider:
The numerator is found by adding the above cases.