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Given P(A|B∪C)≥min(P(A|B),P(A|C)), prove or disprove the statement. My intuition said that the statement should be correct, however I can't think of a mathematical way of proving this statement. Any suggestion is highly appreciated

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    If you $a$re **given** P(A|B∪C)≥min(P(A|B),P(A|C)), what is there to prove or disprove?2012-02-20

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Try

A B C Prob 0 1 0 0.2 0 0 1 0.2 1 1 1 0.6 

Then $\Pr(A|B)=\Pr(A|C)=0.6/0.8 = 0.75$

but $\Pr(A|B \cup C) = 0.6/1.0 = 0.6$.

So the statement is not always true.

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    @TearGrant: don't forget to 1. upvote an answer (an uparrow to the left of the answer) if you like it and 2. accept it (check sign, also to the left)2012-02-22