What is the probability of obtaining a poker hand(5 cards)
containing 2 aces?

We can choose the pair of aces any of 4C2 ways.
For each of these choices of a pair of aces we can choose the three
non-aces any of 48C3 ways.

So the numerator of the desired probability is (4C2)(48C3)

The denominator of the desired probability is 52C5

So the desired probability is

(4C2)(48C3)
-----------
   52C5

[(4·3)/(2·1)]·[(48·47·46)/(3·2·1)]
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   (52·51·50·49·48)/(5·4·3·2·1)

= .0399298181