Proving Conditional Probabilities

Date: 01/24/2001 at 20:09:10
From: Mike Walker
Subject: Conditional Probability

Dr. Math, 

I was hoping that you could give me a hand with a short proof. 
(Note: ^c means complement.) 

If P(B|A) > P(B), show that P(B^c|A) < P(B^c).  

I got a hint from my professor as well. He suggested adding P(B^c|A) 
to both sides of the given inequality.  

Any help with this proof would be greatly appreciated.  Thanks in 
advance.

MiKe

Date: 01/24/2001 at 20:19:52
From: Doctor Schwa
Subject: Re: Conditional Probability

Hi Mike,

First let's do it intuitively, then see if we can make it rigorous.

The first inequality says "If A happens, B is more likely to happen
than if it were to happen on its own."

The second inequality says "If A happens, not-B is less likely to
happen than if it were on its own."

It sure seems like a tautology that if the first one is true, the
second one must be too.

Now down to definitions.

P(B|A) = P(A and B) / P(A), so the first inequality can be rewritten 
P(A and B) > P(A) P(B), while the second inequality can be rewritten
P(A and B^c) < P(A) P(B^c).

Since it's not immediately obvious to me how to turn one into the
other, I'll try your hint: add P(B^c | A) to both sides.

Now we have P(B^c | A) + P(B | A) > P(B) + P(B^c | A).

Rewriting the conditionals in terms of the definition,
P(B^c and A)/P(A) + P(B and A)/P(A) > P(B) + P(B^c and A)/P(A).

Multiplying by P(A),
P(B^c and A) + P(B and A) > P(A) P(B) + P(B^c and A).

Now the left side is just the probability that A happens somehow.
It can happen with B, or without ...
P(A) > P(A) P(B) + P(B^c and A)

That is,
P(A) - P(A) P(B) > P(B^c and A)
P(A) (1 - P(B)) > P(B^c and A)
P(A) P(B^c) > P(B^c and A)

Wait a minute, isn't that the second inequality I was trying to prove?
I didn't mean to solve the whole problem for you... I was just messing 
around, using your professor's hint and the definitions of conditional 
probability and complement ...

I guess that's a good general strategy for proving things, taking a 
look at the definitions of the key terms in the proof.

Personally I like the intuitive proof better. I'm left more convinced 
by it than by the half-page of algebra.

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/