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Conditional Probability Question

  1. Jun 8, 2013 #1
    Suppose that there are 2 apple trees. Tree A and tree B.

    A produces 0.7 of the farm's apples. And B produces 0.3.

    Out of the apples that tree A produces, 0.15 are bad. For B, 0.05 are bad.

    One package of goodies contains 3 apples.

    Given this information, what is the P(Tree A| at least one bad apple in the package)

    Okay, so here's what I tried.

    Probability of at least 1 bad apple in the package from tree A is:
    x=0.7*(1-(0.85)^3)
    Probability of at least 1 bad apple in the package from tree B is:
    y=0.3*(1-(0.95)^3)

    Then P(Tree A| at least one bad apple in the package) = x/(x+y) = 0.863

    Does this seem right? Any help would be greatly appreciated.
     
  2. jcsd
  3. Jun 9, 2013 #2

    mathman

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    Science Advisor
    Gold Member

    An exact, but kludgy approach would be using a multinomial (4) distribution.
    A = prob from tree A and good = .595 (no. = j)
    a = prob from tree A and bad = .105 (no. = k)
    B = prob from tree B and good = .285 (no. = m)
    b = prob from tree B and bad = .015 (no. = n)

    term = {3!/(j!k!m!n!)}AjakBmbn where j,k,m,n ≥ 0 and j+k+m+n = 3

    Desired probability is fraction.
    Numerator = sum of terms with k > 0 (contains a bad apple from A).
    Denominator = sum of terms with k > 0 or n > 0 (contains a bad apple).
     
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