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  1. #1
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    Show that...

    P(A U B U C) = P(A) + P(B) + (C)+ - P(A intersect B) - P(B intersect C) - P (C intersect A) + P (A intersect B intersect C).

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  2. #2
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    Quote Originally Posted by bhuang View Post
    P(A U B U C) = P(A) + P(B) + (C)+ - P(A intersect B) - P(B intersect C) - P (C intersect A) + P (A intersect B intersect C).

    I don't know how to show this...
    I assume you are familiar with the identity \mathbb{P}(A \cup B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A \cap B).

    Now
    \mathbb{P}((A \cup B) \cup C)=\mathbb{P}(A\cup B)+\mathbb{P}(C)-\mathbb{P}( (A \cup B) \cap C).

    Expand this out again using the fact that (A \cup B)\cap C=(A \cap C) \cup (B \cap C).
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