Show that if $P(E | F) \geq P(G | F)$ and $P(E | F^c) \geq P(G | F^c)$ then $P(E) \geq P(G)$

Question

Problem statement: Show that if $P(E | F) \geq P(G | F)$ and $P(E | F^c) \geq P(G | F^c)$ then $P(E) \geq P(G)$

Logically, this makes very much sense to me since $F \cup F^c = S$

This leads me to my attempt:

Since $F \cup F^c = S$ we can say that $P(E | F) \cup P(E | F^c) = P(E)$ (same with $G$). And therefore we can substitute: $P(E) \geq P(G)$

However, is $P(E | F) \cup P(E | F^c) = P(E)$ and actual thing? And if yes, how would you proof it more formal and detailed?

Answer 1

$$\begin{align} P(E)=P(E\cap F)+P(E\cap F^c)&=P(E|F)P(F)+P(E|F^c)P(F^c)\\ \\ &\ge P(G|F)P(F)+P(G|F^c)P(F^c)\\ \\ &=P(G\cap F)+P(G\cap F^c)=P(G) \end{align}$$

Answer 2

To answer your remaining question: The expression $P(E|F)\cup P(E|F^c)$ doesn't make sense, since $P(E|F)$ and $P(E|F^c)$ are numbers. The operator $\cup$ works on sets (events), not numbers.

You probably meant to argue $$P(E|F) + P(E|F^c) = P(E).$$ This isn't true either, because (for instance) if $E$ and $F$ are independent events, then the LHS simplifies to $P(E) + P(E)$. So this line of argument doesn't work -- the assertion isn't an actual thing.