Given events A, B, C in a probability space, I want to prove: min(P(C|A), P(C|B))/2 <= P(C|(A or B)) However it doesn't seem to be working out for me, although it makes sense intuitively. Any hints?
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Here it is: Suppose for instance that $\displaystyle P(A)\geq P(B)$. Then $\displaystyle P(A\cup B)\leq 2P(A)$. In addition, $\displaystyle P(C\cap(A\cup B))\geq P(C\cap A)$. These two inequalities should suffice to prove your inequality. Laurent.
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