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Thread: Proof using Kolmogorovs Axioms

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    Re: Proof using Kolmogorovs Axioms

    Quote Originally Posted by elijah123 View Post
    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.
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