# Conditional probability intuition

Regarding this question and its highest-voted answer: Conditional probability intuition.

So I get the idea of using Venn-Diagrams and limiting our "universe" to a new subset, but what happens if we try looking at it with a tree diagram?
Using this equation: $$P(A\vert B)P(B)=P(B\vert A)P(A)$$ we can set the probabilities to whatever we'd like, for example let's say $P(A)=\dfrac12$, $P(B)=\dfrac13$, $P(A|B)=\dfrac14$, $P(B|A)=\dfrac15$.
The equation does not hold, the events are dependent on each other.
If sequence of occurrence does matter, how can you continue using a Venn-Diagram to prove this equality?

You can't set probabilities to whatever you like. $P(A|B)$ is defined as being $P(A,B)/P(B)$ and $P(B|A)$ is defined as being $P(A,B)/P(A)$. If you have $P(A)=1/2$ and $P(B)=1/3$ and $P(A|B)=1/4$ then $P(A,B)=1/12$ and $P(B|A)=1/6$. You can't set $P(B|A)=1/5$ any more than you can set $2+2=5$.
There is a problem with the probabilities that you have assigned: $$\mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} \implies \mathbb{P}(A\cap B) = \frac{1}{12}$$ On the other hand, $$\mathbb{P}(B|A) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)} \implies \mathbb{P}(A\cap B) = \frac{1}{10}$$ You can set the probabilities whatever you would like, but once you have specified any 3 of the following: $\mathbb{P}(A)$, $\mathbb{P}(B)$, $\mathbb{P}(A|B)$, $\mathbb{P}(B|A)$ or $\mathbb{P}(A \cap B)$ the remaining two will be determined.
The proof of the equality is a simple corollary from the definition of conditional probability: $\mathbb{P}(A|B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$ and hence it follows that the occurrence of events should not matter.