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.