What is the probability that a hand has exactly two hearts given that it has the 2 of hearts? . A card game is played with a deck whose cards can be one of 6 suits, one of the suits being hearts, and one of 11 ranks. A hand is a subset of 3 cards. What is the probability that a hand has exactly two hearts given that it has the 2 of hearts? Please explain.
 A: There are $\binom{65}{2}$ hands with the two of hearts, all equally likely. (I'm assuming the deck has 66 cards - one for every suit/rank pair.)
Of those, how many hands have the two of hearts, another heart, and a non-heart?
A: Since there are 6*11 = 66 Cards total in the deck, you can think of this problem as a sequence. What is the probability that someone draws 1 card that is a heart and then draws one card that is a heart and the draws a card that is not a heart. This can be thought of as a sequence of independent experiments with different parameters.
In other words there are initially 11 hearts out of 66 cards So the probability of choosing 1 heart is 11/66
For the second part of the sequence, you will choose another heart, but there are now only 10 hearts remaining since you choose one already. Thus the probability of the second card being a heart is 10/65.
For the last part of the sequence, you will choose a card that is not a heart. Only hearts have been removed from the deck, so there are still 11 ranks * 5 suits that are not hearts in the deck. There are 55 non-hearts remaining. However, there are only 64 cards left in the deck because of the two hearts that were removed. Thus the probability of choosing a non-heart for this card will be equal to 55/64
You can now multiply the 3 probabilities to get the union of the three scenarios. you will find that the probability of choosing a heart, then a heart, then a non-heart is equal to (11/66) * (10/65) * (55/64) = 55/2496. 
However, since you were asked the probability of drawing exactly 2 hearts. You must consider the possibilities that the three cards could be drawn in any order. With this in mind, you can multiply (55/2496) by 3 because there are 3 different ways that 2 Hearts and Non-heart can be chosen.
Observe; HHN, HNH, and NHH
The probability of choosing 2 hearts exactly is 55/2496 * 3 = 55/832.
The probability of choosing 2 hearts given that the two of hearts was chosen can be thought of using independent scenarios as well. 
Since the first sequence is given(ie probability that the one card will be a 2 of hearts), the probability will be equal to 66/66 or 1.
For the second scenario, consider the probability that the second card will be a heart. The probability will be equal to 10/65.
For the final part of the sequence, the probability will be 9/64.
Th probability that the first card will be a heart, then the second card will be a heart, then the next card will not be a heart is equal to (66/66) * (10/65) * (9/64) = 9/416.
There are 2 different ways that the  cards can be chosen given that 1 is a 2 of hearts. Since the two of hearts is a given, there are only two ways that the last two cards can be picked. That is HN and NH. 
Thus, the probability of choosing 2 hearts exactly given that one of the cards chosen is the 2 of hearts is 9/416 * 2 = 9/208
