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Math Help - P(A) = P(B) = 0 prove: PA(UB) = 0

  1. #1
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    P(A) = P(B) = 0 prove: PA(UB) = 0

    Hey people. Im sitting here with an assignment.

    ______________________________

    It is given that P(A) = P(B) = 0
    Then i have to prove that P(A U B) = 0.
    ______________________________

    We are supposed to use the theorem 3.1:
    Let P be a probability measure. Then we have that

    P(A\B) = P(A) - P(B) if A contains B
    and
    P(Ac) = 1 - P(A) where Ac is A complement.

    furthermore, P is increasing in the sense that

    if A contains B it implies that P(A) is greater than or equal to P(B)

    and P is continuous in the sense that An (arrow up) A => P(An) (arrow up) P(A) and the same the other way around.
    _____________________________________

    I hope it makes sense and i hope someone can help me.
    Its a fairly simple thing, but somehow i cant wrap my head around it.
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  2. #2
    MHF Contributor matheagle's Avatar
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    P(aub)=p(a)+p(b)-p(ab)=0+0-0=0 done.
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  3. #3
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    How can you say that P(aub) = P(a)+P(b)-P(ab)?.. in my book the theorem says that the correct equation is
    P(AUB) = P(A) + P(B) - P(A n B) Where n is the upside down U. As far as i know P(AB) isnt equal to P(AB)?
    Can you elaborate?

    Thanks for the help.
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  4. #4
    MHF Contributor matheagle's Avatar
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    what does this mean.............
    'As far as i know P(AB) isnt equal to P(AB)?'
    so x\ne x, very interesting, but..............

    AB means intersection and AB is contained in A , also B
    So it has probability zero
    anyhow you still would have
    P(AUB)=-P(AB)which is less than or eqal to zero...... giving you 0=0.

    It's 3am and I don't have time for this.
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  5. #5
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    heh sorry a typing error from my side. I meant that P(AB) wasnt equal to P(AnB).

    so your saying that P(AB) = P(AnB)? apparently i missed that.
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  6. #6
    MHF Contributor matheagle's Avatar
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    yes, it's just notation

    if you prefer

    P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B)=0+0-0=zero

    SEE (1) on http://www.weibull.com/SystemRelWeb/...ity_theory.htm
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  7. #7
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    Alright, thanks a lot for the help .
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