Probability of getting 3 aces from a full pack without replacement

Question

I am practicing on some probability exercises based on some old notes.
I have the following:
What is the probability of getting 3 aces in a three-fold draw from a full pack without replacement.
My approach was
4 aces and 52 cards so $\frac{4}{52}$ the possibility to get an ace in the first draw and $\frac{3}{52}$ in the second draw and $\frac{2}{52}$ in the last hence:
$\frac{4}{52}* \frac{3}{52}* \frac{2}{52} = \frac{24}{140698} = \frac{3}{17,576}$

but the solution says $\frac{1}{5525}$
What am I doing wrong?

Answer 1

The question states without replacement. Hence the probability is $$ \frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}=\frac{1}{5525}. $$