Bayes’ formula conditions I was just wondering if given Bayes’ formula, 
$P(A|B) = \frac{ P(B|A) P(A)}{P(B)}$
can one always claim that $0 < \frac{P(B|A)}{P(B)} < 1$?
Can a proof be given? 
Note: A and B are different events. 
Thanks. 
 A: No, because it's not true.
Suppose I flip a coin. Let A = "the coin is heads". Let B be the same event. Then $P(B) = 0.5$, but $P(B | A) = 1$, because in the event the coin is heads, the coin is definitely still heads. Thus, $\frac{P(B | A)}{P(B)} = 2$.
Similarly, if B = "the coin is tails", then $P(B | A) = 0$, so $\frac{P(B | A)}{P(B)} = 0$.
A: 
Can a proof be given? 

No, it is not always so.   Many counterexamples exist.
We just need any events $A,B$ where outcomes for event $A$ are underrepresented within event $B$ than within the sample space. 
Let $\Omega =\{1,2,3,4,5,6\}$ be the outcome set for a sample space with no bias between outcomes.   Let our events of interest be: $B=\{1,2,3\}$ and $A=\{2,4,6\}$.   Ie: roll a fair six sided dice, and call $B$ the event of rolling $3$ or less and $A$ the event of rolling an even result.$$\begin{split}\mathsf P(A)&=1/2\\\mathsf P(A\mid B)&=1/3\\\mathsf P(B)&=1/2\\\mathsf P(B\mid A)&=1/3\\[2ex]\frac{\mathsf P(B\mid A)}{\mathsf P(B)}&=\frac 32\end{split}$$
