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.


Two cards are drawn without replacement from a standard deck of playing cards.  Given that at least one ace is drawn, find the probability that the two cards are of different suits.

Method 1: We correct your attempt.
As you observed, the probability that at least one ace is drawn is
$$\Pr(\text{at least one ace is drawn}) = \frac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51} = \frac{33}{221}$$
If we draw two aces, they must be of different suits.  This event can occur in $4 \cdot 3$ ways.
If we draw one of the four aces with the first draw, there are $51 - 3 - 12 = 36$ cards left in the deck which are not aces and are of a different suit than the ace.  Hence, the number of ways of drawing an ace and then drawing a non-ace of a different suit is $4 \cdot 36$.
If we draw one of the $48$ non-aces with the first draw, there are $3$ aces left in the deck with a different suit than the non-ace we drew first.  Hence, the number of ways of drawing a non-ace and then drawing an ace of a different suit is $48 \cdot 3$.
Hence, the probability that the two cards are of different suits and at least one ace is drawn is
$$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51} = \frac{25}{221}$$
Hence, the probability that the two cards are of different suits given that at least one ace is drawn is
\begin{align*}
\Pr(\text{of different suits} \mid & \text{at least one ace is drawn})\\
\qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\
\qquad & = \frac{\dfrac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{52 \cdot 51}}{\dfrac{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}{52 \cdot 51}}\\
\qquad & = \frac{4 \cdot 3 + 4 \cdot 36 + 48 \cdot 3}{4 \cdot 3 + 4 \cdot 48 + 48 \cdot 4}\\
\qquad & = \frac{25}{33}
\end{align*}
Method 2: We work with combinations to avoid considering the order in which the cards are drawn, which simplifies the calculations.
The probability of drawing at least one ace when two cards are drawn is
$$\Pr(\text{at least one ace}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}$$
since we must either select two of the four aces or one of the four aces and one of the $48$ non-aces while selecting two of the $52$ cards in the deck.
The probability that the two cards are of different suits and at least one ace is drawn is
$$\Pr(\text{of different suits} \cap \text{at least one ace is drawn}) = \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}} = \frac{25}{221}$$
since either two aces are drawn or one ace and one of the $36$ non-aces that are of a different suit than the ace are drawn when two cards are drawn from the deck of $52$ cards.
The probability of drawing two cards of different suits given that at least one ace has been selected is
\begin{align*}
\Pr(\text{of different suits} & \mid \text{at least one ace is drawn})\\
\qquad & = \frac{\Pr(\text{of different suits} \cap \text{at least one ace is drawn})}{\Pr(\text{at least one ace is drawn})}\\
\qquad & = \frac{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{52}{2}}}{\frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}{\dbinom{52}{2}}}\\
\qquad & =  \frac{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{36}{1}}{\dbinom{4}{2} + \dbinom{4}{1}\dbinom{48}{1}}\\
\qquad & = \frac{25}{33}
\end{align*}
Note:  Observe that in both methods, after we cancel the common denominators, the numerator reduces to the number of cases in which at least one ace is drawn and cards are of different suits and the denominator reduces to the number of cases in which at least one ace is drawn.  This is because the sample space for the conditional probability that the cards are of different suits given at least one ace is drawn is the set of cases in which at least one ace is drawn.
