# conditional probability of an ace being drawn and then a king

I am trying to study for my statistic exam and not sure how to solve this question. The question is :

what is the possibility of the following event: an ace is drawn first and a king is drawn second.

Here is what I have done but not sure if its right.

Using the conditional probability formula : $$P(A|B) = P(A \cap B)/P(B).$$

so A = ace is drawn ==> 4/52 ==> 1/13 and B = king is drawn ==> 1/13. so now I am not sure how to do $$P(B \cap A)$$ could someone direct me from this point on as to what I need to do to solve this?

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Are you drawing with replacement or without replacement? – user21436 Mar 1 '12 at 4:40
without replacement... – Ahoura Ghotbi Mar 1 '12 at 4:41
By the way, the title should not be "Conditional probability $\dots$." We are just finding the probability. I gave a solution that used conditional probabilities, but it was the probability that second is King given that first is Ace. – André Nicolas Mar 1 '12 at 5:39

The probability that the first card drawn is an Ace is $\frac{4}{52}$. Given that an ace was drawn first, there are $51$ cards left, so the (conditional) probability that a King is drawn next is $\frac{4}{51}$. Thus our required probability is $$\frac{4}{52}\cdot\frac{4}{51}.$$

We can use more machinery. Let $A$ be the event an Ace was drawn first, and let $B$ be the probability that a King is drawn second. We want $P(A\cap B)$. By the usual formula $$P(A\cap B)=P(B|A)P(A).$$ We have $P(A)=\frac{4}{52}$ and $P(B|A)=\frac{4}{51}$. Multiply. Note that this is the same solution as the first one!

Another way: Record the result of the first two drawings as an ordered pair $(X,Y)$. For example, $X$ could be "Jack of $\spadesuit$," and $Y$ could be "$2$ of $\clubsuit$."

There are $(52)(51)$ such ordered pairs, all equally likely.

How many have the shape $(U,V)$ where $U$ is one of the $4$ Aces, and $V$ is one of the $4$ Kings? Clearly $(4)(4)$. So our probability is $$\frac{(4)(4)}{(52)(51)}.$$

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Thanks, very helpful, another quick question. I have read that you can not rewrite (A and B) as P(A) * P(B) unless they are independent events, and in my case, the cards are dependent events and I noticed that you use P(A) * P(B). could you maybe explain this whole dependent and independent thing a bit more for me? – Ahoura Ghotbi Mar 1 '12 at 5:10
If you look at the answer again, you will see that I did not use $P(A)P(B)$. I used $P(A)P(B|A)$. The unconditional probability of $B$, called $P(B)$, is the probability that the second card is a King, given no information about the result of the first draw. It turns out that $P(B)=\frac{4}{52}$. This fact was not used in the calculation. Explaining the whole dependent/independent thing would take a long time. But one can say that $A$ and $B$ are independent if $P(A\cap B)=P(A)P(B)$. First card Ace, second card King are not independent. – André Nicolas Mar 1 '12 at 5:26
again thanks alot for your time. that helped alot and cleared things up. – Ahoura Ghotbi Mar 1 '12 at 5:31
I posted another question here math.stackexchange.com/questions/115176/… Would appreciate it if you could take a look. – Ahoura Ghotbi Mar 1 '12 at 5:50