Discussion
Let \(A, B, C\) be three events such that \(P(C)P(C^{c}) \gt 0\). Consider the following statements.
S1: \(P(A|C) \gt P(B|C)\) and \(P(A|C^{c}) \gt P(B|C^{c})\) together imply that \(P(A) \gt P(B)\).
S2: \(P(A|C) = P(A|C^{c})\) implies that \(A\) and \(C\) are independent.
Then
Answer: C
Both S1 and S2 are correct.
Both follow from the theorem of total probability, which applies because \(P(C)\) and \(P(C^{c})\) are both positive.
S1:
\[ \begin{aligned} P(A) &= P(A|C)P(C) + P(A|C^{c})P(C^{c}) \\ &\gt P(B|C)P(C) + P(B|C^{c})P(C^{c}) = P(B) \end{aligned} \]
The strict inequality survives because each weight is positive. (Simpson's paradox concerns comparisons where the weights differ between the two groups; here both events are weighted by the same \(P(C)\) and \(P(C^{c})\).)
S2: if \(P(A|C) = P(A|C^{c}) = p\), then \(P(A) = p\,P(C) + p\,P(C^{c}) = p\), so \(P(A|C) = P(A)\), which is independence of \(A\) and \(C\).
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