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Probablity Question

  1. Feb 25, 2009 #1
    Hi, I need some help with this question.

    Seventy percent of all Big Burger chain stores decided to advertise in their local newspapers. Of those chain stores that advertised in their local newspapers, 60% had an increase in sales. Of those chain stores that did not advertise in their local newspapers, 25% had an increase in sales.
    (a) What is the probability that a randomly selected store has an increase in sales.
    (b) What is the probability that a randomly selected store with an increase in sales advertised in its local newspaper?

    for a)
    I did
    (.7)(.6) + (.3)(.25) = 0.495

    for b)
    A = Increased sales
    B = Advertised

    P(A|B) = 0.495/0.7 = 0.707


    Did I do this right?
     
  2. jcsd
  3. Feb 26, 2009 #2

    jambaugh

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    Gold Member

    a.) Looks right.
    b.) It looks like you have A & B reversed.
    The question asks for the probability one advertised given one had an increase in sales. So that should be P(B|A) as you defined them.
     
  4. Feb 26, 2009 #3
    So for b), would

    P(B|A) = (.7)(.6)/(.495) = 0.848

    be correct?
     
  5. Feb 27, 2009 #4
    yes it is
     
  6. Feb 27, 2009 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Another way to do (b) is to imagine that there are 1000 stores. Then 70% of them, or 700, advertise. Of those, 60%, or 420, have an increase in sales. 300 stores do not advertise and 25% of them, 75, also have an increase in sales. So a total of 420+ 75= 495 stores have an increase in sales and 420 advertised: that is, 420/495= .848 (approximately) or 84.8% of the stores that had an increase in sales advertised.
     
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