Question 1174895
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I appreciate you showing your work and thought process. 


The denominator is incorrect because you won't be using C(52,2). Those terms cancel when you divide those two fractions. 


Also, we won't use C(12,2). Instead, we use C(6,2) because there are only 6 red cards that are also face cards. 
(3 red face cards per suit)*(2 red suits) = 6 face cards that are red
So we'll ignore cards like the queen of spades. 


Recall from the conditional probability formula
P(A given B) = P(A and B)/P(B)
This means the numerator must involve events A and B occurring simultaneously.

In other words, P(A and B) = C(6,2)/C(52,2)


So we get the following:


A = card is a face card
B = card is red
*[Tex \Large P(\text{A given B}) = P(\text{A and B}) \div P(\text{B})]


*[Tex \Large P(\text{A given B}) = \frac{C(6,2)}{C(52,2)} \div \frac{C(26,2)}{C(52,2)}]


*[Tex \Large P(\text{A given B}) = \frac{C(6,2)}{C(52,2)} \times \frac{C(52,2)}{C(26,2)}] 


*[Tex \Large P(\text{A given B}) = \frac{C(6,2)*C(52,2)}{C(52,2)*C(26,2)}]


*[Tex \Large P(\text{A given B}) = \frac{C(6,2)}{C(26,2)}] ...... The C(52,2) terms cancel


*[Tex \Large P(\text{A given B}) = \frac{15}{325}]


*[Tex \Large P(\text{A given B}) = \frac{3*5}{65*5}]


*[Tex \Large P(\text{A given B}) = \frac{3}{65}]


I skipped steps showing how to compute the nCr combination values of C(6,2) and C(26,2). Let me know if you need me to go over them. 


The <font color=red>final answer in fraction form is 3/65</font>
The numbers 3 and 65 don't have any factors in common (other than 1) so the fraction cannot be reduced further.


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Here's another approach:


We know the cards are both red, so why not just focus on the red cards only? 


We have 26 red cards. Of those red cards, 6 of them are face cards.


We have C(6,2) = 15  ways of pulling out two face cards that are red
We have C(26,2) = 325 ways of pulling out two red cards


This immediately leads to 15/325 = <font color=red>3/65</font> found earlier.


Side note: 3/65 = 0.04615384615384 approximately
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