# Thread: Proof using Kolmogorovs Axioms

1. ## Proof using Kolmogorovs Axioms

Hello everyone hope you're all having a good day,

Here's is the statement I am trying to prove:
Prove that P(Ω \ A) = 1 − P(A), for any Ω, P and A, indicating where Kolmogorov’s axioms are needed.

From what I have thought of so far, P(Ω \ A) = P(Ω) - P(A), by using Axiom II which says P(Ω) = 1, we can deduce that
P(Ω \ A) = P(Ω) - P(A) = 1 - P(A)

My issue is that, I am not sure if the step " P(Ω \ A) = P(Ω) - P(A)" is a correct approach to the question, since it is assuming that it is true without proving it, and no other theorem explicitly states it.
Any help would be appreciated.

Thank you

2. ## Re: Proof using Kolmogorovs Axioms

$\Omega=A\cup\Omega\setminus A$ and this is a disjoint union. So $1=P(\Omega)=P(\Omega\setminus A)+P(A)$. Done.

3. ## Re: Proof using Kolmogorovs Axioms

Originally Posted by elijah123
Hello everyone hope you're all having a good day,

Here's is the statement I am trying to prove:
Prove that P(Ω \ A) = 1 − P(A), for any Ω, P and A, indicating where Kolmogorov’s axioms are needed.
From what I have thought of so far, P(Ω \ A) = P(Ω) - P(A), by using Axiom II which says P(Ω) = 1, we can deduce that
P(Ω \ A) = P(Ω) - P(A) = 1 - P(A)
Here is a fairly well done webpage.