We're drawing 3 cards from a deck of 52, and we have
![\dbinom{52}3=(52!)/(3!(52-3)!)](https://img.qammunity.org/2019/formulas/mathematics/college/969osg8ehr48rqi7jo9aenyl1m7oow5ng5.png)
ways of drawing any 3-card hand.
Of the 4 total aces in the deck, we want to draw 2. The third card can be any 1 of the 48 remaining cards. We have
![\dbinom42\cdot\dbinom{48}1=(4!)/(2!(4-2)!)\cdot(48!)/(1!(48-1)!)](https://img.qammunity.org/2019/formulas/mathematics/college/6fq6afl0wf89z473zbcmjhq9e7pn763smi.png)
possible 3-card hands that contain any 2 aces.
The probability of drawing such a hand is then
![\frac{\binom42\cdot\binom{48}1}{\binom{52}3}=(72)/(5525)](https://img.qammunity.org/2019/formulas/mathematics/college/37d4orn3vzfxs3t5ceiqif08yyvfg8nk56.png)