A jar contains 7 balls, 3 red, 4 white. 2 balls are drawn without replacement, what is the probability of the first ball was white given the second is white?
So I have conditional probability $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$
I get probability the second ball is white as the the probability of having red then white or the probability of white then white. As $\frac{3}{7}\cdot\frac{4}{6}=\frac{2}{7}$ and $\frac{4}{7}\cdot\frac{3}{6}=\frac{2}{7}$, then probability the second is white is $\frac{2}{7}+\frac{2}{7}=\frac{4}{7}$
But I don't know how to use the formula.
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$\begingroup$I think there is some confusion over what an event is and how they are related, so let me try to clarify that in a few different ways.
In this particular problem, we have that:
Event $A$: drawing the first ball, it is white
Event $B$: drawing the second ball, it is white
Picture each event as an action with a result, and this is how we can assign probabilities to them. We say that these two particular probabilities are dependent if the probability of $B$ happening can change depending on whether or not $A$ happens, and vice-versa.
This, however, is not the same as saying $P(B)$ depends on $P(A)$. The two values $P(A)$ and $P(B)$ are fixed because they are referring to the total probability, beginning from before any ball is drawn.
Try to picture the events and possible ball draws in the following Venn diagram:
We can let the left circle represent $A$ and the right circle represent $B$. The middle represents $A \cap B$, or drawing two white balls. The outside of the Venn diagram represents $\neg A \cap \neg B$, where we draw two red balls.
$P(A)$ is the entirely of the set of the left circle divided by all possible outcomes, including the ones outside the Venn diagram ($\neg A \cap \neg B$). $P(A \mid B)$ means we don't even look outside the domain of the set $B$. Within only the right circle, what is the set of $A$ (in this case, the intersection) divided by the set of $B$?
Drawing red ($\neg A$) then drawing white ($B$) gives us the region outside the left circle but inside the right circle. Drawing white ($A$) then white ($B$) gives us the region inside both circles. When you add these regions, you get the probability of the entire right circle, $P(B)$.
Additional notes and shortcuts for solving this problem:
The problem is symmetrical between the first and second draws of the balls. This is because just as we can draw ball $1$ and then ball $2$, we can also draw ball $2$ and then ball $1$. Therefore, $P(A \mid B) = P(B \mid A)$. Thus, $P(A \mid B)$ is simply the probability of drawing a single white ball after one white ball is removed, $\frac{3}{6} = \frac{1}{2}$.
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