A ⊂ B ⇒ P (A) ≤ P (B) using result of theAxiomsofProbability How can I proof

P(B) ≤ P (AUB ) and

P (AB)≤ P(A) ?

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- February 16th 2010, 11:56 AMwadasA ⊂ 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) ?* - February 16th 2010, 12:06 PMicemanfan
If A represents a subset of the events corresponding to B, then if A occurs, B occurs.

- February 16th 2010, 12:25 PMMoo
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... - February 16th 2010, 12:32 PMjohanS
If you want to prove in a more formal way (instead of using the suggestion above) how can you write as a union of disjoint sets? And more easily, what is if ?

Because if you have X and Y disjoint, i.e. ten by the axioms of probability you know that (You will also need to use the fact that for any event .) If your not used to work with sets you can try by using venn diagrams. - February 16th 2010, 12:34 PMjohanS
sorry, when I started writing your reply wasn't here

- February 16th 2010, 12:55 PMicemanfan
- February 16th 2010, 01:01 PMMoo
Yes, but it can't be taken as a formal proof of what's being said here (Nod)

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 :( - February 16th 2010, 01:18 PMicemanfan