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Proving P(B)>=P(A) using Kolmogrov's axioms

  1. May 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Suppose A is a subset of B. Using the three axioms to establish P(B)≥P(A)


    2. Relevant equations
    Axioms...
    1. For any event A, P(A)[itex]\geq[/itex]0
    2.P(S)=1
    3.If events A1,A2... are mutually exclusive, then P([itex]\cup[/itex]Ai)=[itex]\sum[/itex]P(Ai)


    3. The attempt at a solution

    From the venn diagram I drew, i have...

    B=A+[itex]\overline{A}[/itex]B
    P(B)=P(A[itex]\cup[/itex][itex]\overline{A}[/itex]B)
    P(B)=P(A)+P([itex]\overline{A}[/itex]B)

    According to axiom 1, P([itex]\overline{A}[/itex]B)≥0 ... but dosen't that mean i just proved P(B)≤P(A)?
     
  2. jcsd
  3. May 8, 2014 #2

    jbunniii

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    Does your notation [itex]\overline{A}[/itex]B mean the relative complement of ##A## in ##B##, in other words ##B \setminus A = B \cap A^c##?
    If so, then I agree with your proof.
     
  4. May 8, 2014 #3
    Yes, that is what i meant. Maybe the solution is wrong then... i guess i'll have to ask my professor.
     
  5. May 8, 2014 #4

    jbunniii

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    Sorry, I misread your question:
    No, you proved ##P(B) \geq P(A)##, just like the question asked. Your obtained this result:
    $$P(B) = P(A) + P(B \setminus A)$$
    which means that ##P(B) \geq P(A)##, since you have to add the nonnegative number ##P(B \setminus A)## to ##P(A)## in order to obtain ##P(B)##.
     
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