Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

"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.

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.