Probability: $A,B,C\subseteq \Omega$ (where $\Omega$ is the probability space.) If $ A \subseteq B$, Prove or disprove $P(A) \le P(A\mid B)$ A,B,C are three events in our probability space
Two questions: 
1.) $A,B,C\subseteq \Omega$ (where $\Omega$  is the probability space.) If $ A \subseteq B$, Prove or disprove $P(A) \le  P(A\mid B)$
2.)  $A,B,C\subseteq \Omega$ (where $\Omega$  is the probability space.) If $ B \subseteq A$, Prove or disprove $P(A) \le  P(A\mid B)$
This is my train of thought:
1.) If A is a subset of B, then at most A will be equal to B (is this correct?) or smaller than B. Let's assume that A is equal to B, then we have (where P is the probability): 
$$P(A\mid B) = \frac{P(A \cap B)}{P(B)} $$ 
Now because A at most the space probability as B, then $P(A \cap B) =0$ (Is this correct as well?) and we get that it's false. 
Question two is very similar to question one, so if my though process is correct then I should be able to handle it.
I appreciate any help as my foundations for statistics and probability are quite shaky.

Hint: $P(A | B) = \dfrac{P(A \cap B)}{P(B)}$.
If $A \subseteq B$, what does the above formula reduce to?  If $B \subseteq A$, what does the above formula reduce to?
