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A 2nd year stat/probability question about P(A|B) I'm sooo close

  1. Feb 11, 2008 #1
    A 2nd year stat/probability question about P(A|B) I'm sooo close!!

    1. The problem statement, all variables and given/known data
    Consider 2 events A and B such that P(A)=x and P(B)=y, given x,y >0 and x + y > 1

    Prove that
    P(A|B) >= 1 - [(1-x)/y]


    2. Relevant equations

    P(A|B) = P(A n B) / P(B)


    3. The attempt at a solution

    Now I know that P(A)= x so P(not a) = 1-x and P(B) = y and P(Not B) = 1-y

    so the answer that I have to prove is actually the same as 1 - P(not a)/P(b)

    So all I have to work out is how to go from

    P(A|B) = P(A n B) / P(B)

    to

    1 - P(not a)/P(b)

    And I'm going off the part where x + y > 1

    What step am I missing? Am I using the right formula? There must be a trick or formula I can manipulate as this is a question at the start of a chapter so should be easyish..?
     
  2. jcsd
  3. Feb 11, 2008 #2
    [tex]1\geq P(A\cup B)=P(A)+P(B)-P(A\cap B)[/tex]

    [tex]1\geq x+y-P(A\cap B)[/tex]

    [tex]P(A\cap B)\geq x+y-1[/tex]

    [tex]\frac{P(A\cap B)}{P(B)}\geq \frac{x+y-1}{y}[/tex]

    [tex]P(A|B)\geq \frac{y-(1-x)}{y}[/tex]

    [tex]P(A|B)\geq 1-\frac{1-x}{y}[/tex]
     
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