Show that, if $P(A\mid C)> P(B\mid C)$ and $P(A\mid C^c)> P(B\mid C^c)$, then $P(A)>P(B)$.

Question

Any help with this question would be appreciated. I have trouble with anything related to $p(a\mid b)$.

Answer 1

HINT: Either $C$ or $C^c$ happens. Conditioning on whether $C$ or $C^c$ happens, we can see that $$\mathbb{P}(A) = \mathbb{P}(A ~|~ C) \mathbb{P}(C) + \mathbb{P}(A ~|~ C^c) \mathbb{P}(C^c) \\ \mathbb{P}(B) = \mathbb{P}(B ~|~ C) \mathbb{P}(C) + \mathbb{P}(B ~|~ C^c) \mathbb{P}(C^c)$$ These expressions are symmetric. But whether or not $C$ happens, event $A$ happens with higher probability than event $B$ (i.e. both of the conditional probabilities are higher for $A$ than for $B$). What does this say about $\mathbb{P}(A)$ compared to $\mathbb{P}(B)$?