Conditional probability greater than unconditional probability I am trying to derive conditions for where $P(A|B) > P(A)$,
By Bayes' theorem, I get that:
$P(B|A) P(A) / P(B) > P(A)$, 
which means that I am left with: $P(B|A) > P(B)$. 
What would the interperetation of this result be? 
Thanks. 
 A: Well, the correlation coefficient is $\rho_{\lower{0.5ex}{A,B}}=\dfrac{\mathsf P(A\cap B)-\mathsf P(A)\,\mathsf P(B)}{\sqrt{\bbox[0.5ex]{\mathsf P(A)\,(1-\mathsf P(A))\,\mathsf P(B)\,(1-\mathsf P(B))}}}$ .
So the condition is met when there is a positive correlation between the random variables.
A: As Graham Kemp commented, just use
$$P(A\mid B) = \frac{P(A\cap B)}{P(B)}
$$
to get $P(A\cap B) > P(A)P(B)$. There is no need (or benefit) to appealing to Bayes' theorem here.
A: Here is a simple condition directly derived from the definition of conditional probability. You want this:
$$P(A|B)=\frac{P(A\cap B)}{P(B)} \stackrel{!}{>} P(A)$$
But fact is that $A\cap B \subseteq A$. So,
$$P(A|B)=\frac{P(A\cap B)}{P(B)} \leq \frac{P(A)}{P(B)} \mbox{  and  } \frac{P(A)}{P(B)} \gt P(A) \mbox{  for  }0 \lt  P(B) \lt 1$$
So, a working condition follows easily:
$$A \subseteq B \mbox{ and } P(B) <1$$
A: Suppose that after observing Alice for an entire year, including both rainy days and dry days, we observe that she carried an umbrella on $15\%$ of the occasions when she left the house.
But we also observe that Alice took an umbrella $90\%$ of the time when it was raining.
We estimate that if $B$ is the event that Alice carries an umbrella the next time she leaves her house, and $A$ is the event that it is raining then,
then $P(B) = 0.15$ and $P(B\mid A) = 0.8.$
Alice just left the house carrying her umbrella. Is it more likely than usual for it to be raining right now?
