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

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Given that P(A|C) > P(B|C) and P(A|[tex]\bar{C}[/tex]) > P(B|[tex]\bar{C}[/tex]), 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. jcsd
  3. Oct 15, 2007 #2

    CompuChip

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    Homework Helper

    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
    [tex]P(A \mid B) = \frac{P(A \cap B)}{P(B)}[/tex]

    Not guaranteed to work, but it might.
     
  4. Oct 15, 2007 #3

    Gib Z

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    Remember that [itex]P(A or B) = P(A) + P(B) - P(A and B)[/itex].
     
  5. Oct 29, 2007 #4
    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!
     
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