How to find the probability of drawing a certain two cards from a pack Could I please ask for help on the last part of this question:

Two cards are drawn without replacement form a pack of playing cards. Calculate the probability:
a) That both cards are aces
b) that one (and only one) card is an ace
c) That the two cards are of different suits
d) Given that at least one ace is drawn, find the probability that the two cards are of different suits.

Here's my attempt (for parts a, b, and c I get the answer given in the book):
Let:
$A =$ Event that both cards are aces
$B =$ Event that one and only one card is an ace
$C =$ Event that the two cards are of different suits.
a) $P(A) = \frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$ (as must pick an ace AND another ace)
b) $P(B) = \frac{4}{52} \cdot \frac{48}{51} + \frac{48}{52} \cdot \frac{4}{51} = \frac{32}{221}$ (as can pick ace then not ace, or not ace than ace)
c) $P(C) = \frac{13}{52} \cdot \frac{39}{51} \cdot 4 = \frac{13}{17}$ (as can pick any given suit first, followed by not that same suit, and this can be done in four ways, one for each suit).
d) Let $D =$ Event that at least one ace is drawn.
$P(D) = P(A) + P(B)$ (because at least one ace is drawn only "if both cards are aces" or "one and only one card is an ace")
so $P(D) = \frac{1}{221} + \frac{32}{221} = \frac{33}{221}$
Now, I need to calculate $P(C \mid D) = \frac{P(C \cap D)}{P(D)}$
So if I can calculate $P(C \cap D)$ then I can divide this by $P(D)$ to get the answer.
I (wrongly it appears!) reasoned like so:
To end up with two cards where "at least one is an ace and both are of different suits" you can only have this by either having "the first card be an ace and the second a card of a different suit from that ace" OR having "the first card be of a certain suit and the second card an ace of another suit".
Let @ stand for any suit.
so
\begin{align*}
P(C \cap D) & = P(\text{ace of @}) \cdot P(\text{not @}) + P(\text{@}) \cdot P(\text{ace not of @})\\
            & = \frac{4}{52} \cdot \frac{39}{51}  \cdot 4 + \frac{13}{52} \cdot \frac{3}{51} \cdot 4\\
            & = \frac{5}{17}
\end{align*}
Well this leads to $P(C \mid D) = \frac{5}{17} \cdot \frac{221}{33} = \frac{65}{33}$.
Answer given in book is $\frac{25}{33}$.
Thanks for any help.

Here's how I would think about these:
a) Initially there are 52 cards in the deck, 4 of them aces.  The probability the first card you draw is and ace is 5/52= 1/13.  Now there are 51 cards in the deck, 3 of them aces.  The probability the second card is an ace is 3/51= 1/17.  The probability the two cards are both aces is (1/13)(1/17)= 1/221.
b)  Initially there are 52 cards in the deck, 4 of them aces.  The probability the first card you draw is and ace is 5/52= 1/13.  Now there are 51 cards in the deck, 51- 3= 48 of them not aces.  The probability the second card is NOT an ace is 48/51= 16/17.  The probability the  two cards are an ace and not an ace, in that order is (1/13)(16/17)= 16/221.  The probability the first card you draw is NOT an ace is 48/52= 12/13.  Now there are 51 cards in the deck, 4 of them aces.  The probability the second card is an ace is 4/51.  The probability the two cards are not an ace and an ace, in that order, is (12/13)(4/51)= (4/13)(4/17)= 16/221.   The probability of an ace and not an ace, in either order, is 16/221+ 16/221= 32/221.
c)  The first card MUST be of some suit.  There are then 51 cards left, 52- 13= 39 (or 3*13= 39) of a different suit from the first.  The probability the second card is of a different suit from the first is 39/51= 13/17.  The probability the two cards are of different suits is 13/17.
d)  The probability the two cards are  both aces is, as in (a), 1/221, and, in that case, they must be of different suits.  If one of the cards is an ace there are 52- 4= 48 "non-aces" and 3*12= 36 of them are not of the same suit as the ace.  The probability of that is 36/48= 3/4. The probability, given that at least one card is an ace, that they are of different suits is 1/221+ 3/4= 1/884+ 664/884= 665/884.
