# 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!