Given an urn with $6$ red and $8$ white balls, $4$ balls are selected after each other without replacement.
I want to calculate the conditional probability that two red and two white balls are drawn (event $B$) if the first one is white and the last one is red (event $A$).
Obviously, there are $\binom{6}{2}\binom{8}{2}$ possibilities to draw two red and two white balls and there are $\binom{14}{4}$ total possibilities to draw four balls. This means that $$\mathbb{P}(B)=\frac{\binom{6}{2}\binom{8}{2}}{\binom{14}{4}}$$
Now for event $A$: The probability that the first ball is white is $8/14$, but for the last one being red, this is difficult to say.
My attempt: Knowing that we only need one more white and one more red ball, I argue that $\mathbb{P}(\text{second ball white, third red})=\frac{7}{13}\frac{6}{12}$ and $\mathbb{P}(\text{second ball red, third white})=\frac{6}{13}\frac{7}{12}$, so the total probability becomes $2\frac{6}{13}\frac{7}{12}=\frac{7}{13}$
How do I solve this problem using the formal definition $\mathbb{P}(B|A)=\mathbb{P}(A\cap B)\mathbb{P}(A)^{-1}$?
You can actually just pretend that you have no information about the second and third draws. Then
$$ P(A) = \frac{8}{14} \times \frac{6}{13} = \frac{24}{91} $$
However, I'm not certain that doing this helps you that much, because now you need to obtain $P(A \cap B)$, and the easiest way to do that in this particular problem, is to compute $P(B \mid A)$, and then $P(A \cap B) = P(B \mid A)P(A)$. But there's no point in doing that anymore, because you'll already have $P(B \mid A)$.
Suppose you've drawn four balls (say, without looking at them). Now you look at the first and last ball, and they are in fact white and red, respectively. (That is, event $A$ has occurred.) For event $B$ to have occurred, you need the middle two balls to be one white, one red, in some order. Therefore,
\begin{align} P(B \mid A) & = \frac{7}{12} \times \frac{5}{11} + \frac{5}{12} \times \frac{7}{11} \\ & = \frac{35}{132} + \frac{35}{132} \\ & = \frac{35}{66} \end{align}