# Math Help - A ⊂ b ⇒ p (a) ≤ p (b)

1. ## A ⊂ b ⇒ p (a) ≤ p (b)

A ⊂ B ⇒ P (A) ≤ P (B) using result of the Axioms of Probability How can I proof
P(B) ≤ P (AUB ) and
P (AB)
≤ P(A) ?

2. If A represents a subset of the events corresponding to B, then if A occurs, B occurs.

3. Hello,

It's not a matter of events, since we want to use the axioms, what you said is not valid ^^'

To prove that A ⊂ B ⇒ P (A) ≤ P (B), just consider the disjoint sets BnA and BnA', where A' denotes the complement of A.
Their union makes B, and by the third axiom of probability, you can conclude...

P(B) ≤ P(AUB) comes from the fact that B ⊂ AUB
P (AB) ≤ P(A) comes from the fact that AB ⊂ A

That's all...

4. If you want to prove in a more formal way (instead of using the suggestion above) how can you write $A \cup B$ as a union of disjoint sets? And more easily, what is $A \cup B$ if $A \subset B$ ?
Because if you have X and Y disjoint, i.e. $X \cap Y = \emptyset$ ten by the axioms of probability you know that $P(X \cup Y) = P(X) + P(Y)$ (You will also need to use the fact that for any event $E, P(E) \geq 0$.) If your not used to work with sets you can try by using venn diagrams.

6. Originally Posted by Moo
It's not a matter of events, since we want to use the axioms, what you said is not valid ^^'
Sorry, I was trying to convey the following idea: Let A represent snowing in Chicago and B represent snowing somewhere in Illinois. Clearly, A is a subset of B and if A occurs, then B occurs. Isn't this a universal pattern?

7. Originally Posted by icemanfan
Sorry, I was trying to convey the following idea: Let A represent snowing in Chicago and B represent snowing somewhere in Illinois. Clearly, A is a subset of B and if A occurs, then B occurs. Isn't this a universal pattern?
Yes, but it can't be taken as a formal proof of what's being said here

Why would A be a subset of B, though they're events ? There's a degree of abstract that can't be found if we talk about "if event A occurs, then so does B". It's a mere translation of the inequality between the probabilities that there are in the first message.

I don't know if I explained well what I thought, I hope you understand what I want to say

8. Originally Posted by Moo
I don't know if I explained well what I thought, I hope you understand what I want to say
I think I understand what you are saying. I could elaborate on my point, but you have aptly demonstrated how to solve the problem, so there really is no need for that.

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