From a pack of 52 cards, two cards are drawn in succession one by one...

There are $52$ cards in a pack of cards.
There are $4$ aces in that pack.
Event of choosing one card $n\left(S\right)=C152$
Event of choosing an ace $n\left(A1\right)=C14$
If one card is drawn at random, then the probability of getting an ace is $P\left(A1\right)$
$\Rightarrow$$\frac{n\left(A1\right)}{n\left(S\right)}=\frac{C}{1}$ $\left(\because P\left(A\right)=\frac{n\left(A\right)}{n\left(S\right)}\right)$
$=\frac{4}{52}=\frac{1}{13}$
There are $51$ cards in that pack.
There are $3$ aces in that pack.
Event of choosing one card $n\left(S\right)=C151$
Event of choosing a second ace $n\left(A2\right)=C13$
If one card is drawn at random, then the probability of getting a second ace without any replacement is $P\left(A2\right)$
$\Rightarrow$$\frac{n\left(A2\right)}{n\left(S\right)}=\frac{C}{1}$ ($\because$one card is drawn at random)
$=\frac{3}{51}=\frac{1}{17}$
The odds of receiving two aces without any replacement are calculated by calculating the odds of getting an ace and a second ace without any replacement.
$P\left(A1\right)\times P\left(A2\right)=\frac{1}{13}\times \frac{1}{17}=\frac{1}{221}$