"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.
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– 2012-12-14