Probability of at most two aces I am dealt 7 cards from a standard pack. I want to find the probability of being dealt at most two aces by using combinations.  I believe the calculation looks like this:
$$\frac{\binom{4}{2} \binom{48}{5}+\binom{4}{1} \binom{48}{6} +\binom{4}{0} \binom{48}{7}}{\binom{52}{7}}=\frac{7689}{7735}=0.994053005817... .$$ 
Is this the correct solution? Thanks

Community wiki answer so the question can be marked as answered:
As has been pointed out in comments, this calculation (as corrected) is correct.
