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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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    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.2012-03-01

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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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    I posted another question here http://math.stackexchange.com/questions/115176/a-probability-question-using-percentages Would appreciate it if you could take a look.2012-03-01