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Math Help - [solved] p(a)=1

  1. #1
    Moo
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    [solved] p(a)=1

    Hi

    A little question...

    Let \Omega be the probability space.
    We know that if P(A)=1, then P(\Omega \backslash A)=0 (we don't necessarily have A=\Omega)

    Is there a proof (with formulae, not diagrams) for that :
    For any X \subset \Omega, if P(A)=1, then P(A \cap X)=P(X)
    ?

    Thanks
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    P(A \cap X)=P(X \mid A) P(A)
    if P(A)=1 its pretty clear right?
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hi

    A little question...

    Let \Omega be the probability space.
    We know that if P(A)=1, then P(\Omega \backslash A)=0 (we don't necessarily have A=\Omega)

    Is there a proof (with formulae, not diagrams) for that :
    For any X \subset \Omega, if P(A)=1, then P(A \cap X)=P(X)
    ?

    Thanks
    Here's my proof:
    P(X)=P(A\cap X)+P(A^c\cap X) where A^c=\Omega\setminus A,
    and 0\leq P(A^c\cap X)\leq P(A^c)=1-P(A)=0,
    so that P(X)=P(A\cap X).

    It has the significant advantage (compared to the previous one) of avoiding to divide by 0.
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  4. #4
    Moo
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    Quote Originally Posted by Robert Hall View Post
    Since

    P(A \cap X)=P(X \mid A) P(A)
    if P(A)=1 its pretty clear right?
    Hmm So how would you prove that P(X \mid A)=P(X) ? That's the problem...

    Quote Originally Posted by Laurent View Post
    Here's my proof:
    P(X)=P(A\cap X)+P(A^c\cap X) where A^c=\Omega\setminus A,
    and 0\leq P(A^c\cap X)\leq P(A^c)=1-P(A)=0,
    so that P(X)=P(A\cap X).

    It has the significant advantage (compared to the previous one) of avoiding to divide by 0.
    That's... erm perfect... :P

    Too bad Plato deleted his post, I can't remember it completely ><


    Thank you to all three for replying
    And special thanks to Laurent, because he can always provide the answer I'm looking for
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