Why is the numerator of Baye's Theorem $P(A\cap B)$ instead of $P(A|B)$?

Question

For example, say I have am holding an object; $O$ is the event that the object is an orange, and $R$ is the event that the object is round.

$P(O|R) = \frac{P(R|O)}{P(R)}$

This is obviously incorrect, but it reads to me like: "the probability that the object is an orange given that it is round is the probability of an object being round given that it is an orange, divided by the probability that the object is round in general". Which seems reasonable.

Why is the above interpretation wrong?

Answer 1

“Philosophically” it’s wrong because the right hand side is about roundness — the only place Orange appears is as a given condition — so not sure how you can get any info on Orange on LHS from only data about roundness on RHS.

Conceptually, it’s just wrong.

From the definition of conditional probability:

$$P(O|R)= \frac{P(O\cap R)}{P(R)}$$

However:

$$ \frac{P(R|O)}{P(R)}= \frac{\frac{P(O\cap R)}{P(O)}}{P(R)}=\frac{P(R\cap O)}{P(R)P(O)}\neq P(O|R)$$