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Probability Proof

  1. Aug 29, 2012 #1
    We have three events A, B, and C such that
    P(A|C)[itex]\geq[/itex]P(B|C) and P(A|C')[itex]\geq[/itex]P(B|C')

    Prove that P(A)[itex]\geq[/itex]P(B).

    First I started with,
    P(A|C)=P(AC)/P(C) and similarly P(B|C)=P(BC)/P(C).

    From above,
    P(AC)/P(C)[itex]\geq[/itex]P(BC)/P(C) and since P(C)[itex]\geq[/itex]0,
    P(AC)[itex]\geq[/itex]P(BC).

    Similarly from above,
    P(AC')[itex]\geq[/itex]P(BC').

    Here is where I am stuck. I tried to expand P(AC)=P(A)+P(C)-P(AUC) with each piece above, but I didn't seem to get anywhere. Any ideas would be greatly appreciated.
     
    Last edited: Aug 30, 2012
  2. jcsd
  3. Aug 29, 2012 #2

    vela

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    How did you get P(A|C)=P(AB)/P(C)? Why is B there?
     
  4. Aug 30, 2012 #3
    Typo, fixed it above.
     
  5. Aug 30, 2012 #4
    If I'm not mistaken then P(A and C) + P(A and not C) = P(A).
     
  6. Aug 30, 2012 #5

    HallsofIvy

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    You have this backwards. It should be [itex]P(A\cup C)= P(A)+ P(C)- P(A\cap C)[/itex]

     
  7. Aug 31, 2012 #6
    Okay, just figured it out. Thanks everyone for the assistance.
     
    Last edited: Aug 31, 2012
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