"You are dealt 13 cards randomly from a pack of 52. What is the probability your hand contains exactly 2 aces?"
I thought about breaking it down into:
${4 \choose 2}$ = number of ways to choose two of four aces.
${48 \choose 11}$ = number of ways to choose 11 cards from the non-aces.
${52 \choose 13}$ = choose any 13 cards from the 52.
And then using:
$\cfrac{\binom{4}{2} \cdot \binom{48}{11}}{\binom{52}{13}}$
Could anyone confirm this solution or show otherwise?
Best Answer
Community wiki answer so the question can be marked as answered:
As noted in the comments, your solution is correct.