Baye's Theorem, conditional probability question You are dealt three cards. The events of interest concern the number of face cards that you are dealt ($0,1,2,$ or $3$).  Construct a tree diagram tracking face cards.

d) What is the conditional probability that you are dealt three face cards given that you are dealt at least two face cards.  

The answer is $.07692$, but I cannot for the life of me figure it out.  I don't see how it's not $0.0095$
 A: As the original poster has not written up the answer, I'll expand it here in case anyone else lands on this question.
In theory, the conditional probability $P(A|B)$, that of event $A$ happening given that event $B$ already occurred, is:
$$
\frac{P(A \cap B)}{P(B)}
$$
In English, "A given B" means we want to know the probability of both $A$ and $B$ happening if $B$ already happened. The probability of both $A$ and $B$ happening is ${P(A \cap B)}$ and the universe is the set of all events in which $B$ has happened, so our denominator is not 1, but $P(B)$.
Specifically, in a standard poker deck of 52 cards (no jokers) there are 12 face cards and 40 non-face cards. The probability we want is the number of ways to get 3 face cards—$P(A \cap B)$—divided by the number of ways to get at least two face cards—$P(B)$.
The number of ways to get three face cards is:
$$
{12 \choose 3}\cdot{40 \choose 0} = 220
$$
The number of ways to get at least two face cards is the number of ways to get two face cards and the number of ways to get three face cards. This is:
$$
{12 \choose 2}\cdot{40 \choose 1} + {12 \choose 3}\cdot{40 \choose 0} = 2640 + 220 = 2860
$$
So the answer is $\frac{220}{2860} \approx 0.076923$ as stated above.
