Is it possible there are A,B,&C such that P(A)<P(B) & at the same time P(A|C)>P(B|C) & P(A|C')>P(B|C')?

Assume the second two parts:





P(A|C) %26gt; P(B|C)


P(A|C') %26gt; P(B|C')





Rewriting them:





P(A∩C) / P(C) %26gt; P(B∩C) / P(C)


P(A∩C') / P(C') %26gt; P(B∩C') / P(C')





Simplifying:





P(A∩C) %26gt; P(B∩C)


P(A∩C') %26gt; P(B∩C')





Adding:





P(A∩C) + P(A∩C') %26gt; P(B∩C) + P(B∩C')





Combining:





P(A) %26gt; P(B)





So if the conditional probabilities are in favor of A, the overall probability must also be in favor of A.

Is it possible there are A,B,%26amp;C such that P(A)%26lt;P(B) %26amp; at the same time P(A|C)%26gt;P(B|C) %26amp; P(A|C')%26gt;P(B|C')?
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