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Math Help - Proof of axioms of conditional probability

  1. #1
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    Proof of axioms of conditional probability

    Show that P(A given B) = P(A intercept B) / P(B) implies the following axioms:

    1.) P(A given B) = 1, where B is a subset of A

    2.) If {Ai intersect B},i = 1 2 are mutually exclusive, then
    P(A1 ∪ A2 ∪ given B)= P(A1 given B)+ P(A2 given B)+

    3.) If B ⊃ H and B ⊃ G and P(G ) ≠ 0 , then
    P(G )/P( H ) = P (G given B)/ P(H given B)
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  2. #2
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    Quote Originally Posted by trukka View Post
    Show that P(A given B) = P(A intercept B) / P(B) implies the following axioms:
    1.) P(A given B) = 1, where B is a subset of A
    2.) If {Ai intersect B},i = 1 2 are mutually exclusive, then
    P(A1 ∪ A2 ∪ given B)= P(A1 given B)+ P(A2 given B)+
    3.) If B ⊃ H and B ⊃ G and P(G ) ≠ 0 , then
    P(G )/P( H ) = P (G given B)/ P(H given B)
    For #1, just notice that B \subseteq A\; \Rightarrow \;B \cap A = B.

    For #2, given that \left( {A_k  \cap B} \right) is a collection of pair-wise disjoints events then P\left[ {\bigcup\limits_k {\left( {A_k  \cap B} \right)} } \right] = \sum\limits_k {P\left( {A_k  \cap B} \right)} .
    Then note that you have a common denominator of P(B).

    #3 is just algebra
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