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Axioms of Probability

  1. Apr 25, 2005 #1
    In my probability class we were given two versions of probability axioms which are equivalent. Let S be the sure event, A and B any arbitrary events, I the impossible event. I will use u to denote union, and n to denote intersection:

    Version 1
    1. P(S)=1
    2. P(A)>=0
    3. If AnB=I, than P(AuB)=P(A)+P(B)

    Version 2
    1. P(S)=1
    2. P(A)>=0
    3. P(I)=0
    4. P(AuB)=P(A)+P(B)-P(AnB)

    It is easy to go from version 2 to version 1, and I can see how to show that P(I)=0 using version 1. The problem I'm having is how to derive axiom 4 of version 2 from version 1. I know I will have to consider two events made up of unions and intersections of A and B, to which I can apply axiom 3 of version 1, but I can't quite figure out how to do it. Thanks for any help.
  2. jcsd
  3. Apr 25, 2005 #2
    After some trial and error I got this: partition A u B into A n ~B, A n B, and B n ~A. If you work that out you can get the right expression; one of the terms you'll get is P(A n B), and there is a way to go from P(A n B) to -P(A n B) and the other terms will fall into place.
  4. Apr 25, 2005 #3
    I tried this but wasn't able to work it out. How do you go from P(AnB) to -P(AnB)? Once I get to the second step I'm not sure how to write the two terms other than P(AnB) in any other way, since they involve intersections, and the axioms version 1 don't say anything about intersections. You can rewrite the other two terms (before you make them into 2 seperate terms) as (An~B)u(Bn~A)=(AuB)n(~Bu~A), but once again this involves intersections, not unions which doesn't help much.
  5. Apr 25, 2005 #4
    Nevermind I got it. You just have to turn P(AnB) into 2P(AnB)-P(AnB), then group one of each of the positive terms with your other two terms. Axiom 3 of version 1 can than be used in reverse to turn each of these into a single probability, where the events in question simplify, using boolean algebra, to A and B. Thanks for the help Bicycle Tree.
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