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