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?
 A: 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
$$\frac{(4)(4)}{(52)(51)}.$$
