Confused where and why inequality sign changes when proving probability inequality "Let A and B be two events in a sample space such that 0 < P(A) < 1. Let A' denote the complement of A. Show that is P(B|A) > P(B), then P(B|A') < P(B)."
This was my proof:
$$ P(B| A) > P(B) \hspace{1cm} \frac{P(B \cap A)}{P(A)} > P(B) $$
$$P(B \cap A) + P(B \cap A') = P(B) \implies P(B \cap A) = P(B) - P(B \cap A') $$
Subbing this into the above equation gives
$$ P(B) - P(B \cap A') > P(B)P(A)  $$
I think the inequality was supposed to change there, but I don't know why. Carrying on with the proo and dividing both sides by P(B) and rearranging gives
$$ 1 - P(A) > \frac{P(B \cap A')}{P(B)} $$
$$ P(A') > \frac{P(B \cap A')}{P(B)} $$
Rearrange to get what you need:
$$ P(B) < \frac{P(B \cap A')}{P(A')} = P(B |A') $$
Why does the inequality change at that point?
EDIT: Figured it out. It's in the last line where the inequality holds.

In general $P(B)=P(A)P(B|A) + P(A')P(B|A')$.  What happens if $P(B|A)>P(B)$ and $P(B|A')\geq P(B)$?
Hint: Use $P(A)+P(A')=1$ and $P(A)>0$ and $P(A')\geq 0$ to get a contradiction.
Your proof was right up to (and including) this step:
$$P(A') > \frac{P(B \cap A')}{P(B)}$$
From here, multiply both sides by $\frac{P(B)}{P(A')}$ and you get:
$$P(B) > \frac{P(B\cap A')}{P(A')} = P(B|A')$$
That was what you wanted to prove.
