Card arrangement probability question

Question

Two cards are randomly chosen from a deck of 52 playing cards. What is the conditional probability they are both aces given that they are of different suits?

I'm so stuck please help.

Answer 1

Let $A$ be the event that both cards are Aces, and let $S$ be the event that the cards are of different suits. By definition, $$\Pr(A|S)=\frac{\Pr(A\cap S)}{\Pr(S)}=\frac{\Pr(A)}{\Pr(S)},$$ because if both cards are Aces, they are certainly of different suits, so that $A\cap S = A$. Now $$\Pr(A)=\frac{\binom42}{\binom{52}2}=\frac{4\cdot3}{52\cdot51}=\frac1{13\cdot17}$$ and $$\Pr(S)=\frac{39}{51}$$ because after the first card is drawn $39$ of the remaining $51$ cards are of a different suit than the first. Therefore, $$\Pr(A|S)=\frac1{13\cdot17}\frac{51}{39}=\boxed{\frac1{169}}$$

Answer 2

There are $4 \cdot 3=12$ ordered ways to draw a pair of aces. How many ordered ways are there to draw two cards of different suits? Then divide.

Answer 3

Given that they are of different suits, it is equivalent to draw one card from suit A (whatever it is, the one that corresponds to the first card) and one card from a different suit B. In each draw you have a probability of $\tfrac{1}{13}$ to have an ace so in total the probability is $\left(\tfrac{1}{13}\right)^2=\tfrac{1}{169}$