P(A) given that P(A|B) and P(B) are known Given that $P(A \mid B)$ is known, and $P(B)$ is known, how can $P(A)$ be calculated? I tried using the definition of conditional probability - $P(A \mid B) = P(A \cap B)/ P(B)$ - to solve for $P(A \cap B)$ but I don't know where to go from there. Any hints? 
 A: In general, you can't calculate $P(A)$ from just $P(A \mid B)$ and $P(B)$ — you also need to know $P(A \mid \lnot B)$.
If you do know that, the calculation is easy enough:
$$\begin{aligned}P(A) 
&= P(A \land B) + P(A \land \lnot B) \\
&= P(A \mid B)P(B) + P(A \mid \lnot B)P(\lnot B) \\
&= P(A \mid B)P(B) + P(A \mid \lnot B)(1 - P(B))
\end{aligned}$$
A: You don't have enough information to get $P(A).$
Consider a dice with six faces. Consider $A_1=\{1,2\},A_2=\{2\},B=\{2,4,6\}.$ It is $P(A_i/B)=\frac13$ and $P(B)=\frac{1}{2}.$ However $P(A_1)=\frac13$ and$P(A_2)=\frac16$ 
A: It cannot be.  To see this, let's look at a couple of examples:
First: suppose that you flip a fair coin twice, and let $A$ be the event that the first is heads and $B$ the event that the second is heads.  The coin flips are independent; so, in this case,
$$
P(A)=.5\qquad P(B)=.5\qquad P(A\mid B)=.5
$$
Second: say we again flip two fair coins, and again let $B$ be the event that the second is heads.  However, let $A$ be the event that BOTH are heads. Then
$$
P(A)=.25\qquad P(B)=.5\qquad P(A\mid B)=.5
$$
If there was a way to determine $P(A)$ from $P(B)$ and $P(A\mid B)$, then these two examples would need to have the same value for $P(A)$; but, they don't!
