Given only $P(A)$ and $P(A|B)$ can $P(A \cap B)$ be calculated?

Question

If $P(A)$ and $P(A | B)$ are known, is it possible to calculate $P(B)$ or $P(AB)$?

Using definition of conditional probability: $P(A|B) = P(AB)/P(B)$

Here we don't know $P(AB)$ nor $P(B)$, so two unknowns and $1$ equation is impossible to solve uniquely.

Using Bayes theorem: $P(A|B) = [P(B|A)P(A)]/P(B)$

Here we don't know $P(B|A)$ nor $P(B)$ so again $2$ unknowns.

Answer 1

No, it cannot.

Consider the following examples:

P(A) = 0.1
P(A|B) = 0.2
P(B) = 0.3
P(A and B) = P(A|B) * P(B) = 0.2 * 0.3 = 0.06

vs.

P(A) = 0.1
P(A|B) = 0.2
P(B) = 0.4
P(A and B) = P(A|B) * P(B) = 0.2 * 0.4 = 0.08

The same values of $P(A)$ and $P(A|B)$ yield different values of $P(A \text{ and } B)$, because $P(B)$ can be anything.

Answer 2

I prepared following example: let's take $A=\{ 1,2,3 \}$, $B_1=\{ 1,4\}$ and $B_2=\{ 1,2,4,5\}$ in space $\Omega=\{1,2,3,4,5,6 \}$

Then, fixing $A$ we have $$P(A|B_1)=\frac{1}{2}=P(A|B_2)$$ While $P(B_1) \ne P(B_2)$.