Calculating $P(A\cap B)$ from $P(A)$ and $P(B)$ without knowing $P(A\mid B)$ or $P(B\mid A)$

Question

Is there a way to calculate $P(A\cap B)$ from $P(A)$ and $P(B)$ without knowing $P(A\mid B)$ or $P(B\mid A)$?

I'm asking because it seems to me that a primary function of Bayes' theorem is to calculate $P(B\mid A)$ from $P(A\mid B)$ and $P(B)$ and $P(A)$. However, if one can calculate $P(B\mid A)$ from just $P(B)$ and $P(A)$ without needing $P(A\mid B)$ then Bayes' and indeed the second general multiplication principle would be unnecessary. Is this possible?

I'm talking about conditional probability btw.

Thanks

Answer 1

No, you can't. Only if you knew that $A$ and $B$ are independent, you could write $P(A , B) = P(A) P(B)$. But knowing that is actually also knowing the conditionals, as $A$ and $B$ are independent iff $P(A | B) = P(A)$

In general, $P(A , B)$ has more information than $P(A)$ and $P(B)$ alone: you can obtain the later from the former - eg : $P(A) = \sum_B P(A, B)$ , but not the reverse.