1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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)


    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook