Two cards are drawn from a deck of cards. What is the probability that they are both face cards and at least one is red? Assume that there are $52$ cards and without replacement.
I have two different methods, and they got different solutions. The second method below is the same as the teacher's answer: $0.0385$, but the first is not. What is wrong with my first solution?
Method 1:
$\frac{12\cdot11}{52\cdot51}\cdot\frac{3}{4}=0.0373$. I got this because I calculated the probability that I get two face cards, and then I multiplied by $\frac {3}{4}$ because there is $\frac {3}{4}$ chance that I get at least $1$ red.
Method 2: $\frac{\binom{12}{2}}{\binom{52}{2}}-\frac{\binom{6}{2}}{\binom{52}{2}}=0.0385$. The first fraction is the probability that both are face cards. The second fraction is the probability that they are both face cards and both black.
Best Answer
Your mistake is in thinking that there is a $\tfrac34$ chance to get at least $1$ red; this would be the probability when drawing with replacement. But without replacement, the color of the one card affects the possibilities for the color of the other card.
If you count the number of ways to draw two face cards, you will find there are $\tbinom{12}{2}=66$ ways. The number of ways to draw no red cards is $\binom{6}{2}=15$, so the number of ways to draw at least one red card is $66-15=51$. This means the probability of at least one of the two drawn cards being red is $\tfrac{51}{66}\neq\tfrac{3}{4}$. Your second computation also reflects this.