Finding $P(A \cap B)$, given $P(A\mid B)$ and $P(B\mid A)$ Given events $A$ and $B$, if $P(A\mid B)$ and $P(B\mid A)$ are known, can $P(A \cap B)$ be found? I tried the following approach and came to an answer, but doubt its veracity. Here's my attempt:
I tried thinking of $P(A\mid B)$ as $P(B \implies A)$. This is logically equivalent to $(1-P(A))+P(B)-(1-P(A))P(B)$, right? This comes from $w \implies z = \neg w \lor z$ for Boolean variables. Then it treats $X$ and $Y$ as independent variables, and uses the common formula for the probability of event one or event two happening, $P(A\cup B) = P(A)+P(A)-P(A)P(B)$. In addition, it uses $P(\neg A)=1-P(A)$.
Taking that approach, if you assign $P(A\mid B)=a$ and $P(B\mid A)=b$, this system of equations results:
$$
1-P(A)+P(B)-(1-P(A))P(B)  = a
$$
$$
1-P(B)+P(A)-(1-P(B))P(A)  = b
$$
Well, this can be solved uniquely for $P(A)$ and $P(B)$. Given that solving this system with substitution involves solving a quadratic, the result is
$$
P(A\cap B) = \frac{3}{2} a + \frac{3}{2} b - 2 + \frac{1}{2}(a-b)^2
$$
OR
$$
P(A\cap B) = -\frac{1}{2} a - \frac{1}{2} b + 1 - \frac{1}{2}(a-b)^2
$$
Which looks dubious, since the answer here seems like it should be unique. Moreover, one but not both of these formulas give negative probability for certain combinations of $a$ and $b$.
Well, I believe applying Bayes' theorem is the right way to solve this problem, but it requires knowing either $P(A)$ or $P(B)$ beforehand. Is there a way to obtain $P(A\cap B)$ without the individual probabilities? In addition, what's wrong with the approach above? I assume it has to do with assuming independence between $A$ and $B$.

Your assumption that $$p(A\mid B)=p(A \implies B)$$ is not valid.
Think of $A$ as $x>3$ and $B$ as $x>5$ 
