Question about implication and probablity Let $A, B$ be two event. My question is as follows:
Will the following relation holds:
$$A \to B \Rightarrow \Pr(A) \le\Pr(B) $$
And why?
 A: In terms of intuition, the fact that some event $A$ implies some event $B$ means that whenever $A$ happens, $B$ happens. But if the event $B$ happens, we might not have event $A$. So in other words, we have that $\mathbb P(A) \le \mathbb P(B)$ because the probability that $A$ happens is also "the probability that $B$ happens because of $A$", which is less than "the probability that $B$ happens" with no constrains on it. (Note that there are cases with equality, for instance when $A \Rightarrow B$ and $B \Rightarrow A$. )
From a theoretical point of view though (this is a little bit more advanced for a beginner course in probability though), the expression "$A \Rightarrow B$" doesn't make much sense, since the way probabilities are defined is that the "$\mathbb P$" is actually a function from something we call an space of possible events to the interval of real numbers $[0,1]$. The possible events are sets, and the right way to say $A \Rightarrow B$ in this system would be that the set $A$ is included in the set $B$, i.e. $A \subseteq B$. (You need to think about those sets as "a regrouping of possibilities", for instance if the event space is all subsets of $\{1,2,3,4,5,6\}$ in the context where we roll a fair dice, an example of event would be the possibility that "the roll is even" or "the roll is a $1$ or a $4$".) Since in the construction of probabilities, one most common axiom is that a probability function is countably additive, or in other words
$$
\mathbb P \left( \bigcup_{i=0}^{\infty} A_i \right) = \sum_{i=0}^{\infty} \,\mathbb P (A_i)
$$
when the sets $A_i$ are pairwise disjoint, and another axiom would be that $P(\varnothing) = 0$, we can deduce from that that
\begin{align}
\mathbb P(B) = \mathbb P( (A \cap B) \cup (A^c \cap B)) 
& = \mathbb P( (A \cap B) \cup (A^C \cap B) \cup \varnothing \cup \dots) \\
& = \mathbb P(A \cap B) + \mathbb P(A^c \cap B) + \mathbb P (\varnothing) + \mathbb P (\varnothing) + \dots \\
& = \mathbb P(A)  + \mathbb P(A^c \cap B) + 0 + 0 + \dots \\
& \ge \mathbb P(A).
\end{align}
Note that the reason I added this is because I used the axiom "countably additive" and not "finitely additive". The way to show that countably additive implies finitely additive is by adding plenty of $\varnothing$'s after the finitely many sets. 
There are many possible axioms you can add/choose that are equivalent though. I just took my favorite.
Hope that helps,
A: Yes. Since events are sets of states, $A\implies B$ means $A\subseteq B$, which is equivalent to $B=A\cup(B\backslash A)$, a disjoint union. So $P(B)=P(A)+P(B\backslash A)\geq P(A)$. 
