Are H1 and H2 independent? Supper you pick three cards, one at a time without replacement, from a pack of 52 cards. Let Hi be the event that the ith card is hearts.
What's the probability of H1 conditioned on exactly two of the cards being hearts?
For this I was confused about what I would be putting for the condition because it doesn't specify the two heart combinations. My initial thought was to have it be something like this P(H1 | (H2 and H3)) but I don't think that makes sense but picking H2 and H3 is dependent on H1. I am having issues with specifying the condition.
Are the events H1 and H2 independent?
For this I tried using this formula P(H1 and H2) = P(H1)P(H2), where P(H1) = 13/52 and P(H2) is 12/51. I don't know how to find P(H1 and H2) since this tree has a height of three.
 A: We do a tree calculation. Let us find the probability that we have exactly two hearts. There are $3$ ways that this can happen: HHN (heart, heart, not heart), HNH, and NHH.
The probability of HHN is $\frac{13}{52}\cdot\frac{12}{51}\cdot \frac{39}{50}$. Call this number $p$. Compute the other two probabilities. Each turns out to be $p$. (This is not an accident!)
In exactly two if these cases, the event $H_1$ happens. So our required probability is $\frac{2p}{3p}$, which is $\frac{2}{3}$. 
For the second question, it should be intuitively clear that $H_1$ and $H_2$ are not independent: If the first card is a heart, that makes a heart on the second card less likely than if the first card is not a heart.
But we can also compute. We have $\Pr(H_1\,\,\text{and}\,\, H_2)=\frac{13}{52}\cdot \frac{12}{51}$. 
Now compute $\Pr(H_1)$ and $\Pr(H_2)$. It is clear that $\Pr(H_1)=\frac{13}{52}$. It should be intuitively clear that $\Pr(H_2)$ is also $\frac{13}{52}$, but may not be.
I will try to persuade you that $\Pr(H_2)=\frac{13}{52}$. The intuition is that all sequences of $3$ cards are equally likely. But if this is not persuasive enough, we can do a tree calculation. The event $H_2$ can happen in two ways: (i) HH and (ii) NH. 
The probability of HH is $\frac{13}{52}\cdot\frac{12}{51}$. The probability of NH is $\frac{39}{52}\cdot \frac{13}{51}$. Add and simplify. We end up with $\frac{13}{52}$.
Finally, we conclude from the computation that $\Pr(H_1\,\,\text{and}\,\, H_2)\ne \Pr(H_1)\Pr(H_2)$. 
