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Probability of defective machines question

  1. May 20, 2009 #1
    [solved]probability question

    1. The problem statement, all variables and given/known data

    Three machines A, B, and C produce 50%, 30% and 20% respectively of the total
    number of items at a factory. Each produces a number of defective: 3%, 4% and 5%
    respectively.
    (a) If an item produced by one of these machines is selected, find the probability that
    it is defective.
    (b) Now suppose that a manufactured item is selected at random and is found to be
    defective. Find the probability that this item was produced by machine A.

    2. Relevant equations

    N/A

    3. The attempt at a solution

    is part (a)
    Pr = (3%+4%+5%)/(50%*30%*20%)?

    and no idea on part b

    thanks a lot!
     
    Last edited: May 21, 2009
  2. jcsd
  3. May 20, 2009 #2
    Use Bayes' theorem.

    Let D be the event that a selected item is defective

    Let A, B, and C be the events that that an item is produced by machine A, B, and C respectively

    [tex] P(D) = P(D\cap A) + P(D \cap B) +P(D \cap C) [/tex]

    [tex] = P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C) [/tex]

    [tex] P(A|D) = \frac {P(A \cap D)}{P(D)}[/tex]

    [tex] = \frac {P(D|A)P(A)}{P(D|A)P(A) + P(D|B)P(B) + P(D|C)P(C)} [/tex]
     
  4. May 21, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    How I would do it: Imagine that 1000 of the items are produced. Then 50% of 1000= 500 come from machine A, 30% of 1000= 300 from machine B, and 20% of 1000= 200 from machine C.

    Of the 500 from machine A, 3% of 500= 15 are defective, 4% of 300= 12 are from machine B, and 2% of 200= 4 are from machine C.

    A total of 15+ 12+ 4= 41 defective items are produced of a total of 1000 items. With that information (a) is easy.

    Of a total of 41 defective items, 15 of them were from A. With that information, (b) is easy.
     
  5. May 21, 2009 #4
    thank you guys!!
     
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