# Question about conditional probability?

The rule for conditional probability gives that

$P (A\mid B)P (B)=P (B\mid A)P (A)=P (A \cap B)$

When I am considering the first two expressions, I am confused. Why should they be equal? If event B occurring changes the probability of event A occurring, then I just don't understand why they would necessarily change in such a way that the probability of A occurring after B (multiplied by the probability of B) would necessarily be the same as the probability of B occurring after A (multiplied by P (A)).

I just cannot understand how the changes in probabilities of A and B must be related in that way.

Any help in understanding this (examples and developing intuition would be great) would be much appreciated. I really want to develop a better intuitive understanding of why some mathematical statements must be true, not just by churning through the mathematics and seeing that it leads to this result.

Additional note: I have tried to think about this in terms of Venn diagrams, however I do not see how this can tell us anything. When I consider a Venn diagram, I think of it showing the probabilities of A and B occurring at any given point in time. Now of B occurs, then the probability of A occurring would change too. So surely the Venn diagram would change and the Venn diagram drawn before does not tell us anything about the situation now?

## 2 Answers

Let's take a look at a Venn diagram, masterfully drawn in MS paint:

Let's assume that the area of the whole rectangle is $1$, just to make it easy. In that case, $P(A)$ is the area of the $A$ circle (ellipse?), and $P(B)$ is the area of the $B$ circle. In addition, $P(A\cap B)$ is the area of the wedge overlap between the circles.

Here is the crucial part I think you haven't taken in over yourself, and the reason you asked the question: $P(A\mid B)$ is the ratio between the area of the wedge and that of the $B$ circle: $P(A\mid B) = \frac{P(A\cap B)}{P(B)}$ ("If we know that we're in the $B$ circle, what is the probability that we're also in the $A$ circle?"). If you now take that fraction, and multiply with $P(B)$, you're left with $P(A\cap B)$. The same thing is true for the $A$ circle.

While the relative sizes of the circles and the overlap might change, it is always true that the area of the wedge is equal to the area of either of the circles times the ratio between the area of the wedge and the area of that circle.

We know P(A|B) = $\frac{P(A \cap B)}{P(B)}$

From first expression -

P(A|B)P(B) = $\frac{P(A \cap B)}{P(B)}.P(B)$

= $P(A \cap B)$

Similarly,

P(B|A)P(A) = $\frac{P(A \cap B)}{P(A)}.P(A)$

= $P(A \cap B)$

So its clear all 3 expressions are equal.

First two expressions -

P(A|B)P(B) = P(B|A)P(A)

$\frac{P(A \cap B)}{P(B)}.P(B)$ = $\frac{P(A \cap B)}{P(A)}.P(A)$

$P(A \cap B) = P(A \cap B)$

• Protip: \cap gives $\cap$ (opposite of \cup: $\cup$). Also, \mid gives vertical lines with some room on either side. It looks nicer. – Arthur Jan 14 '17 at 13:20