How to properly solve the following probability problem?

Question

I'm looking for the formalization of the solution to the following problem:

Consider the random experiment which consists of drawing two cards from a fair deck of 52 cards. What is the probability of getting two aces given that the first card was not put back into the deck after the first drawing?

The answer I know lacks of formality. To me, the previous random experiment should have as sample space the set:

$$\Omega =\{(\omega_1, \omega_2): \omega_1\ \textrm{and}\ \omega_2 \ \textrm{are cards from the deck}\}.$$

Then we should define:

$$A_1=\{(\omega_1, \omega_2): \omega_1\ \textrm{is an ace}\},$$

and

$$A_2=\{(\omega_1, \omega_2): \omega_2\ \textrm{is an ace}\}.$$

Then we are looking for the probability of the event:

$$A_1\cap A_2=\{(\omega_1, \omega_2): \omega_1\ \textrm{and}\ \omega_2\ \textrm{are aces}\}.$$

This should come from:

$$P(A_1\cap A_2)=P(A_1)P(A_2|A_1).$$ But how can I compute $P(A_2|A_1)$ using the above formalism?

As a matter of fact, the event $A_1\cap A_2$ does not capture the event ``both cards are ace, given the fact the first card is out of the deck''.

Answer 1

Just use Bayes' Theorem: $\mathsf P(A_2\mid A_1)~=\dfrac{\mathsf P(A_1\cap A_2)}{\mathsf P(A_1)}$

Which should match the intuition:

When given that the first card is an ace, there are 3 aces among the 51 remaining cards that could be drawn second.