How can we conclude about the probability of two events? Suppose there are two events $E_1$ and $E_2$ such that if both $E_1$ and $E_2$ holds then another event $E_3$ that will occur.
How do I derive that $P(E_{1} \cap E_{2}) \leq P(E_{3})$ ?
If $E_{1} \implies E_{2}$ is also true, how can we derive $P(E_{1}) \leq P(E_{2})$ ?
Could anybody explain it with proper proof and example?
 A: The exact proof may differ a bit depending on what definition and properties of probability you're working with, but in broad strokes this is how I would go about it: if $E_1$ happening implies $E_2$ will also happen, then $E_1 \cap E_2 = E_1$: that is to say, $E_1$ occurring is necessary and sufficient to say that $E_1 \cap E_2$ will occur.
So, we can set up two events $E_1 \cap E_2$ and $\overline{E_1} \cap E_2,$ noting that $$(E_1 \cap E_2) \cup (\overline{E_1} \cap E_2) = (E_1 \cup \overline{E_2}) \cap E_2 = E_2$$
$$(E_1 \cap E_2) \cap (\overline{E_1} \cap E_2) = (E_1 \cap \overline{E_1}) \cap E_2  = \emptyset$$
So, we have that
$$\begin{align}\text{Pr}(E_2) & = \text{Pr}(E_1 \cap E_2) + \text{Pr}(\overline{E_1} \cap E_2) \\ & = \text{Pr}(E_1) + \text{Pr}(\overline{E_1} \cap E_2)\end{align}$$
and because we must have $\text{Pr}(\overline{E_1} \cap E_2) \geq 0,$ we must also have $\text{Pr}(E_2) \geq \text{Pr}(E_1).$
As far as the intuition for this goes, I just think about how for every sample point where $E_1$ happens, $E_2$ must also happen, so the set of sample points where $E_2$ will happen is a superset of those where $E_1$ happens, and therefore must be at least as large.
A: *

*

if both $E_1$ and $E_2$ holds then another event $E_3$ that will occur.

In other words, every outcome that is common to events $E_1$ and
$E_2$ also belongs to event $E_3;$ that is, the intersection of
events (sets) $E_1$ and $E_2$ is contained in event $E_3:$ $$E_1\cap
E_2\subseteq E_3.$$ So, $$|E_1\cap E_2|\le |E_3|.$$ Assuming a
finite sample space $S,$ $$\frac{|E_1\cap E_2|}{|S|}\le
\frac{|E_3|}{|S|}.$$ Assuming classical probability, $$P(E_1\cap
E_2)\le P(E_3).$$


*If event $A$ happening implies that event $B$ happens, then every
outcome in event $A$ also belongs to event $B;$ that is, event $A$
is contained in event $B:$ $$A\subseteq B.$$ So, $$A\cap B=A.$$

Now, for arbitrary sets $P$ and $Q,$ set $Q$ a union of the disjoint
sets $P\cap Q$ and $P^\complement\cap Q.$ Thus, $A\cap B$ and
$A^\complement\cap B$ are mutually exclusive and
\begin{align}P(B)&=P(A\cap B)+P(A^\complement\cap
B)\\&=P(A)+P(A^\complement\cap B)\\P(A)&\le P(B),\end{align} noting
that $P(A^\complement\cap B)$ is by definition nonnegative.
