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Prove that if under two conditions that together have probability 1 ($C$ and $C'$), probability of $A$ is greater than probability of $B$, then $P(A) > P(B)$ in general.

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    Do you understand what you ask? – 2017-02-15
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    Use the [total probability rule](http://mathworld.wolfram.com/TotalProbabilityTheorem.html) and it comes out in two steps. If you are still having trouble, post your working and explain where you are stuck, then someone may be able to help further. – 2017-02-15
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    Do you have to use the principle of inclusion-exclusion? I'm unsure what the total is that these parts have to sum to. I'm trying to express it as the sum of intersections: (C ∩ (A U B)) U (C' ∩ (A U B)) = (A ∩ C) U (A ∩ C') U (B ∩ C) U (B ∩ C'). I don't know where to go from there. – 2017-02-15
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    No you do not use the PIE at all. Use the LOTP as David suggested. – 2017-02-15
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    I tried letting the area designated by (A ∩ C) U (A ∩ C') U (B ∩ C) U (B ∩ C') be called E. Then I took probabilities of the portions of E using the LOTP: P(E) = P(A ∩ C) + P(A ∩ C') + P(B∩C) + P(B∩C') P(E) = P(A|C)P(C) + P(A|C')P(C') + P(B|C)P(C) + P(B|C')P(C') P(E) = P(C)(P(A|C) + P(B|C)) + P(C')(P(A|C') + P(B|C')). From here I'm stuck again, since I'm not sure how to relate this to an inequality showing P(A) > P(B). – 2017-02-15
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    I apologize for the lack of new lines; I tried hitting enter and it didn't produce the desired spacing. – 2017-02-15

1 Answers 1

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The LOTP says that $$P(A) = P(A|C) P(C) + P(A|C^\prime)P(C^\prime)$$ and
$$P(B) = P(B|C) P(C) + P(B|C^\prime)P(C^\prime)$$

Since $P(A|C) - P(B|C) >0$ and $P(A|C^\prime) - P(B|C^\prime) >0$, we have $$P(A)-P(B) = \left[ P(A|C) - P(B|C) \right] P(C) +[ P(A|C^\prime) - P(B|C^\prime) ] P(C^\prime) > 0$$