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Question answer I don't understand!

  1. Sep 23, 2013 #1
    An electronic fuse is produced by five production lines in a manufacturing operation.
    The fuses are costly, are quite reliable, and are shipped to suppliers in 100-unit lots.
    Because testing is destructive, most buyers of the fuses test only a small number of
    fuses before deciding to accept or reject lots of incoming fuses.
    All five production lines produce fuses at the same rate and normally produce
    only 2% defective fuses, which are dispersed randomly in the output. Unfortunately,
    production line 1 suffered mechanical difficulty and produced 5% defectives during
    the month of March. This situation became known to the manufacturer after the fuses
    had been shipped.Acustomer received a lot produced in March and tested three fuses.
    One failed. What is the probability that the lot was produced on line 1? What is the
    probability that the lot came from one of the four other lines?

    Let B denote the event that a fuse was drawn from line 1 and let A denote the event
    that a fuse was defective. Then it follows directly that

    P(B) = 0.2 and P(A|B) = 3(.05)(.95)^2 = .135375.

    Similarly,
    P(B-) = 0.8 and P(A|B-) = 3(.02)(.98)2 = .057624.



    P(A) = P(A|B)P(B) + P(A|B-)P(B-)
    = (.135375)(.2) + (.057624)(.8) = .0731742.


    P(B|A) = P(B & A) / P(A) = P(A|B)*P(B) / P(A) = (.135375)(.2) / .0731742 = 0.37

    Wat I don't understand here how did he get those values for
    P(A|B) and P(B-)..? shouldn't P(B | A) = P(A & B) / P(B) = 0.05/0.2

    P(B) = 1/5 = 0.2. same logic for P(A|B-) I don't understand this if someone could explain this more clearly.
     
  2. jcsd
  3. Sep 23, 2013 #2

    Office_Shredder

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    It should be P(A) in the denominator, not P(B) like you have written. You may be confused because the formula will typically be presented as giving P(A | B), so your A and B are flipped from what the standard formula in a textbook would read.
     
  4. Sep 24, 2013 #3
    oh i see
     
  5. Oct 3, 2013 #4

    HallsofIvy

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    Another way of looking at it: Imagine that every line produces 1000 items, for a total of 5000 items. Lines 2 through 4 have 2% bad: a total of .02(4000)= 80 bad items. Line one produces 5% bad, a total of 50 bad items. That is, out of a total of 80+ 50= 130 bad items, 50, or 50/130 or about 38% came from line 1.
     
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