Probability Independence - Determining if two sets are independent (drawing two cards) I've got a few problems here that I feel pretty confident on. I am asking for confirmation on these answers. However, I am stuck on problem #3. Please let me know if you need more information.

Two cards are drawn, one at a time, without replacement, from a deck
  of $52$ cards. Start from a new deck in each problem. Determine if the two sets are independent.

1. Let $A$ be the event that the two cards have the same rank. Let $B$ be the event that the first card is an ace.
I answered yes, $P(A) = \frac{1}{17}$, $P(B) = \frac{1}{13}$, $P(A \cap B) = \frac{1}{221}$, thus $P(A \cap B) = P(A)P(B).$
2. Let $A$ be the event that the two cards have the same rank. Let $C$ be the event that the two cards have the same suit.
I answered no, because $P(A) = \frac{1}{17}$, $P(C) = \frac{4}{17}$, and $P(A \cap B) = 0$, because you can't have two cards in a standard deck of the same rank and suit. $0 \neq P(A)P(B)$.
3. Let $D$ be the event that the first card is a club. Let $E$ be the event that the second card is a club.
I've found that $P(D) = \frac{1}{204}$. However, I don't know how to find $|E|$ or similarly $P(E)$. My intuition says I can't without knowing what my first card was, but I don't know how to write this.
 A: In 3), we do not have independence. At the informal level, this is clear: if the first card is a club, that makes it less likely that the second card is a club.
More formally, let $A$ be the event the first card is a club, and $B$ the event the second card is a club. We have $\Pr(A)=\Pr(B)=\frac{13}{52}$.
The probability both cards are clubs, that is, $\Pr(A\cap B)$, is $\frac{13}{52}\cdot\frac{12}{51}$. This is not equal to $\Pr(A)\Pr(B)$. 
A: Your method looks okay.
For (b): You (correctly) obtained $\tfrac 4{17}$ for two cards having the same suit.   Well then, just use the same logic for finding the union in (c).
So $\mathsf P(D\cap E) = \tfrac 1 4\cdot \tfrac{12}{51} = \tfrac 1 {17}$  -- it is also the probability that two cards will have the same suit and that suit will be the favoured suit.
Now $\mathsf P(D) = \tfrac {13}{52} = \tfrac 1 4$; a choice of selecting $1$ of $13$ ranks of the favoured suit from $52$ cards without bias.   Likewise $\mathsf P(E)=\tfrac 1 4$.   And clearly they aren't going to multiply to give $\tfrac 1{17}$.
