Need explanation of the numerator in Bayes Theorem I've seen plenty of answers explaining the denominator but none for the denominator. I.e:
Numerator:    $P(A|B) = P(B|A) \times P(B)$
What I don't understand is, we already have the conditional probability in $P(B|A)$ why is it we also need to multiply it by $P(B)$?
As I understand it, (and imagining a Venn diagram) $P(B|A)$ is already the intersection of A and B we care about, by multiplying by $P(B)$ aren't we effectively increasing our intersection by the space of $P(B)$?
I hope I've made my doubts somewhat clear. Thank you.
 A: You're conflating $P(B\vert A)$ with $P(B\,\&\,A)$; the latter corresponds to intersection, but the former is more complicated than that.
$P(B\vert A)$ is the probability that $B$ is true given that $A$ is true. Thinking of this in terms of sets, this is a comparison between the intersection $A\cap B$ and $A$ itself. For example, if $B\supseteq A$ then no matter how small $A$ and $B$ each are we will have $P(B\vert A)=1$. Thinking (very coarsely) in terms of cardinalities of sets, $P(B\vert A)$ is the analogue of the ratio $${\vert A\cap B\vert\over \vert A\vert}$$ rather than the outright cardinality $\vert A\cap B\vert$.
At this point, we can write a suggestive "equation:" $${\vert A\cap B\vert\over \vert B\vert}={\vert A\cap B\vert\over \vert A\vert}\cdot{\vert A\vert\over \vert B\vert}$$ which - while ignoring the fact that it doesn't really make sense as soon as we're talking about infinite sets - is the set-flavored analogue of $$P(A\vert B)=P(B\vert A)\cdot {P(A)\over P(B)}$$ (if we want to actually make this rigorous in general we should talk about measure).
