Very simple proof of P(A) > P(B)

1. Oct 14, 2007

ChickenChakuro

1. The problem statement, all variables and given/known data
Given that P(A|C) > P(B|C) and P(A|$$\bar{C}$$) > P(B|$$\bar{C}$$), prove that P(A) > P(B)

2. Relevant equations
How do I write the proof formally?

3. The attempt at a solution
This seems to me intuitively obvious. There are only two sample spaces, either C occurs or does not occur. In both instances, A has a higher chance of ocurrence than B. So how can P(A) NOT be always greater than P(B)? I have been told that P(A) > P(B) can be proven mathematically. However I'm not good at formal proofs. Can anyone help me write this out in mathematical form?

Many thanks!

2. Oct 15, 2007

CompuChip

I'd try something like this:
P(A) = P(B) [ P(A | B) / P(B | A) ]
What can you say about the bracketed term, possibly using that
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$

Not guaranteed to work, but it might.

3. Oct 15, 2007

Gib Z

Remember that $P(A or B) = P(A) + P(B) - P(A and B)$.

4. Oct 29, 2007

dedo

I managed to work this proof down to P(A)-P(AUC)> P(B)-P(BUC) and P(A)-P(AU~C) > P(B)-P(BU~C) but im not sure where to go from there!