prove conditional probability inequality Prove that if $P(A | B) &gt; 0$ then $P ( B | A) \geq P(B)$. I've a couple of different things but I didn't get any closer to solution. Any suggestions would be appreciated.
I know that $ P (B | A) = \frac{P(A \cap B)}{P(A)}$. The fact that $P(A | B) &gt; 0$ basically tells me that $A$ and $B$ intersect somewhere and their intersection is non zero, correct? Sadly I don't have any idea on how to continue.
The inequality seems false indeed let consider for example
$$A=\{1,2,3,4\} \quad B=\{3,4,5,6\}$$
then

$P(A)=\frac23$
$P(B)=\frac23$
$P(A \cap B)=\frac13$
$P (A | B)=P (B | A)=\frac12$
The inequality seems false indeed let consider for example
$$A=\{1,2,3,4\} \quad B=\{3,4,5,6\}$$
then

$P(A)=\frac23$
$P(B)=\frac23$
$P(A \cap B)=\frac13$
$P (A | B)=P (B | A)=\frac12$