I want to prove that knowing $P(A \mid C) > P(B \mid C)$ and $P(A \mid C^c) > P(B \mid C^c)$ then $P(A) > P(B)$. I feel like I know this is true, but every time I try to break down the formulas with Baye's formulas and such I keep hitting a dead end. Is there any way someone can give me a hint on where to start, or maybe this is wrong and there is some counter example I should give.
Once again the idea is to prove that knowing $P(A \mid C) > P(B \mid C)$ and $P(A \mid C^c) > P(B \mid C^c)$ then $P(A) > P(B)$.
 A: Notice that the first relation implies that $\textbf{P}(A\cap C) > \textbf{P}(B\cap C)$.
Similarly, the second restriction implies that $\textbf{P}(A\cap C^{c}) > \textbf{P}(B\cap C^{c})$.
Based on the such assumptions, one has that:
\begin{align*}
\textbf{P}(A\cap C^{c}) - \textbf{P}(B\cap C^{c}) = \textbf{P}(A) - \textbf{P}(A\cap C) - \textbf{P}(B) + \textbf{P}(B\cap C) > 0
\end{align*}
where the last result tells us that:
\begin{align*}
\textbf{P}(A) - \textbf{P}(B) > \textbf{P}(A\cap C) - \textbf{P}(B\cap C) > 0
\end{align*}
which is exactly the property we are looking for.
Hopefully this helps!
A: We know:
\begin{align*}
P(A \mid C) & > P(B \mid C) \\
P(A \mid C^{\mathsf{c}}) & > P(B \mid C^{\mathsf{c}}) \\
P(C) & > 0 \\
P(C^{\mathsf{c}}) & > 0 \\
P(A) & = P(A \mid C) P(C) + P(A \mid C^{\mathsf{c}}) P(C^{\mathsf{c}}) \\
P(B) & = P(B \mid C) P(C) + P(B \mid C^{\mathsf{c}}) P(C^{\mathsf{c}})
\end{align*}
Note that $P(C) > 0$ is an implicit hypothesis, otherwise $P(A \mid C)$ is not defined. Likewise $P(C^{\mathsf{c}}) > 0$ is an implicit hypothesis, otherwise $P(A \mid C^{\mathsf{c}})$ is not defined.
From all these, it follows:
\begin{align*}
P(A \mid C) P(C) + P(A \mid C^{\mathsf{c}}) P(C^{\mathsf{c}}) & > P(B \mid C) P(C) + P(B \mid C^{\mathsf{c}}) P(C^{\mathsf{c}}) \\
P(A) & > P(B)
\end{align*}
