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"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?

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Your reasoning looks fine. –  user84413 Sep 17 '13 at 21:24
Alright, thanks. –  Abhishek M. Sep 17 '13 at 22:36

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