# Probability of Intersections

I'm studying for an exam tomorrow, and I'm definitely over thinking it. Out of a normal deck of $52$ cards, $2$ cards are taken without replacement. Given two events: $A_c$ and $B$, where $A_c =$ {an ace of clubs is chosen} and $B=$ {two aces are chosen}, find $P(B|A_c)$.

What I know is that since there is one ace already selected, the probability of selecting the next ace is $\frac{4-1}{52-1} = \frac{1}{17}$. But if you were to use the formula:

$$P(B|A_c)=\frac{P(A_cB)}{P(A_c)}$$ how would you calculate the intersection of $A_c$ and $B$?

My teacher has that $P(BA_c)= \frac{1}{52}\frac{3}{21}+\frac{3}{52}\frac{1}{51}$. But I'm not sure how he arrived at that.

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Do you mean the intersection of $A_c$ and $B$? –  Ron Gordon Feb 15 '13 at 5:36
@rlgordonma: Yes sorry, fixed now. –  MITjanitor Feb 15 '13 at 5:37

$$P(BA_c)= \frac{1}{52}\frac{3}{51}+\frac{3}{52}\frac{1}{51}$$

That refers to the different ways of getting an ace and an ace of spades. The first term is the ace of spades first, then another ace. The second term is one of the other aces, then the ace of spades.

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Ah yes, a typo in his solutions. But that makes perfect sense. Thank you for the explanation. –  MITjanitor Feb 15 '13 at 5:45
I'd be a little careful with your notation. $A_c$ looks a lot like "$A$ complement." –  Ron Gordon Feb 15 '13 at 5:50

The easy way out of this question is to find the probability of the event where, out of 51 cards (without the $A \clubsuit )$, the other 3 $A$s are selected...which is $\frac {3}{51}$.

Alternatively, with \begin{align*} &A_c \space \text{is the event that A} \clubsuit \space\text{is selected} \\ &B \space \text {is the event that 2 A are selected} \end{align*}

Then

\begin{align*} P(A_cB) &= \frac{3}{52C_2}\\ &= \frac{3}{1326} \text{Probability that 2 As are selected, 1 of which A}\clubsuit\\ P(A_c) &= \frac{51}{1326}\text{Probability that A}\clubsuit\space\text{is selected, along with any other card}\\ P(B|A_c) &= \frac{\frac{3}{1326}}{\frac{51}{1326}}\\ &= \frac{3}{51} \end{align*}

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I know, In my above question, I stated that I knew how to solve it this way. –  MITjanitor Feb 15 '13 at 5:40
Made the necessary changes. –  bryansis2010 Feb 15 '13 at 5:57