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"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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    I think the problem means "If P(B|A)>P(B), then P(B|A').2012-12-14

2 Answers 2

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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.

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Can you think of P(B) as a weighted average between P(B|A) and P(B|A')? How does that help you?