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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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It shouldn't have changed... –  David Mitra Dec 14 '12 at 14:00
    
Why would you think the inequality symbol should be reversed? You do this only when multiplying both sides of an inequality by a negative quantity. If you multiply both sides by a positive quantity, the symbol remains unchanged. –  David Mitra Dec 14 '12 at 14:04
    
I think the problem means "If $P(B|A)>P(B)$, then $P(B|A')<P(B)$. –  Thomas Andrews Dec 14 '12 at 14:07
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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?

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