# Probability Theorem

So I have this theorem which I understand but proving it is more than weird to me. So it is:

If $E_1$ and $E_2$ are subsets of $S$ and $E_1\subseteq E_2$ then $P(E_1)\le P(E_2)$

Which I understand and it is somewhat intuitively logical, but here's the proof:

$E_2=E_1\cup(E_1\cap E_2)$

I understand that, but

$E_1$ is mutually exclusive to $E_1\cap E_2$

Why is it so? If we have two events so that $E_1$ is a subset of $E_2$ then they must have mutual events, so how come is it that their intersection is mutually exclusive to $E_1$?

What I think of is that $E_2$ is equal to what is to union of what is in $E_1$ and what is outside of $E_1$ but in $E_2$ which is absent in $E_1$, but $E_1\cap E_2$ is basically whatever is in both, which to me sounds wrong, is it that the theorem stated above is wrong? (Our universities notes tend to be extremely inaccurate sometimes.)

So my reasoning is that it should be:

$E_2 = E_1 \cup (E_2$ \ $E_1)$

Which is the event 2 is equal to event 1 in union with the relative complement of event 2 and event 1.

• Why do you say that "$E_1$ is mutually exclusive to $E_1\cap E_2$"? It is incorrect.
– JRN
Jan 8, 2014 at 3:20
• "E2 = E1 union (E1 intersection E2)" is incorrect, too. Jan 8, 2014 at 3:23
• @JoelReyesNoche yes of course, here you are: computing.dcu.ie/~jhorgan/chapter5slides.pdf and scroll to the theorem 3 which doesn't even have a name haha... And thank you a lot! Jan 8, 2014 at 3:32
• Oh wow... i opened the notes in the browser and suddenly I can see that it says E1 complement... it's fault of my soda PDF reader. :P Jan 8, 2014 at 3:34
• You are correct. Do you still want me to answer in more detail?
– JRN
Jan 8, 2014 at 3:35

The notes you linked to has the following proof:

$E_2=E_1\cup(\overline{E_1}\cap E_2)$. Now, since $E_1$ and $\overline{E_1}\cap E_2$ are mutually exclusive, $P(E_2)=P(E_1)+P(\overline{E_1}\cap E_2)\ge P(E_1)$ since $P(\overline{E_1}\cap E_2)\ge 0$ from Axiom 1.

As you noticed, your PDF reader omitted the complement on $E_1$.

From your comment, it seems that you want the first statement of the proof clarified. The drawing below shows $\overline{E_1}\cap E_2$. Does it make things clearer now?

• Great! Thank you very very much! You are awesome! :) And thank you for the Venn Diagram :D Jan 8, 2014 at 3:48
• One more thing, is there a name for this theorem? Since the name is missing in the notes :P Jan 8, 2014 at 4:07
• Wikipedia calls it monotonicity (en.wikipedia.org/wiki/Axioms_of_probability#Monotonicity).
– JRN
Jan 8, 2014 at 4:10