"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?

  • 3
    $\begingroup$ Your reasoning looks fine. $\endgroup$ – user84413 Sep 17 '13 at 21:24
  • $\begingroup$ Alright, thanks. $\endgroup$ – Abhishek M. Sep 17 '13 at 22:36

Community wiki answer so the question can be marked as answered:

As noted in the comments, your solution is correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.