Probability that two drawn cards are face cards

Question

Two cards are randomly drawn from the standard deck of cards. Given that the two cards drawn are of different colors(Black and Red), Find the conditional probability that both the cards are face cards.

My try:

Let $A$ be an event of drawing two face cards and $B$ be an event of drawing two cards of different colors.

We need to find $$P\left(A/B\right)$$

By definition we have

$$P\left(A/B\right)=\frac{P(A \cap B)}{P(B)}$$

$$P(A\cap B)=\binom{12}{2}-2\binom{6}{2}=36$$

$$P(B)=\binom{52}{2}-2\binom{26}{2}=676$$

Hence $$P\left(A/B\right)=\frac{36}{676}=\frac{9}{169}$$

But the answer is $\frac{18}{169}$

Anything went wrong?

Answer 1

Another way to look at it is to not consider the problem as a conditional probability problem, but frame it differently. What you are really asking is, given 2 half decks of cards, one with all the red ones, one with all the black ones, what is the chance that I get 2 faces if I draw one card randomly from each of the half decks. So this is simply: $$ \left(\frac{2 \times 3}{26}\right)^2 = \frac{36}{676} $$

I also think that your ground truth answer is wrong.

Answer 2

$$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ with $A$ and $B$ as defined in your question.

$$P(A \cap B) = \frac{\binom{6}{1} \times \binom{6}{1}}{\binom{52}{2}}$$

$$P(B) = \frac{\binom{26}{1} \times \binom{26}{1}}{\binom{52}{2}}$$

Therefore,

$$P(A|B) = \frac{36}{676}$$

The answer given is certainly wrong.