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Math Help - Production line problem

  1. #1
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    Production line problem

    I need to know if I am on the right track with this.

    A factory has 3 shifts. the shifts produce 1%, 2% and 5% defective items respectively. If all shifts have the same productivity, (1) what percentage of the items produced in a day are defective? (2) what is the probability that the defective items were produced on the third shift?

    This is what I came up with:

    let n = total number of items produced daily.
    each shift produces n/3 items.
    we are looking for total (defective/total produced).

    =((n/3)(.01+.02+.05))/(n) which equals .03%

    then...let E denote the probability that a defective item was produced.
    let F denote the probability it was produced on the 3rd shift.

    P(E|F)=P(E n F)/P(F) = .03/.05 = .6 or 60%

    thanks

    Andy
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  2. #2
    MHF Contributor Quick's Avatar
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    Your works correct
    although I haven't checked your arithmetic...
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by andy_atw
    I need to know if I am on the right track with this.

    A factory has 3 shifts. the shifts produce 1%, 2% and 5% defective items respectively. If all shifts have the same productivity, (1) what percentage of the items produced in a day are defective? (2) what is the probability that the defective items were produced on the third shift?

    This is what I came up with:

    let n = total number of items produced daily.
    each shift produces n/3 items.
    we are looking for total (defective/total produced).

    =((n/3)(.01+.02+.05))/(n) which equals .03%
    Check your arithmetic, and look up the difference between a decimal
    fraction and a percentage.

    RonL
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by andy_atw
    then...let E denote the probability that a defective item was produced.
    let F denote the probability it was produced on the 3rd shift.

    P(E|F)=P(E n F)/P(F) = .03/.05 = .6 or 60%
    This part of the question is asking for P(F|E) not P(E|F), and

    P(F|E)=P(E and F)/P(E)=0.3333x0.05/0.03 ~=0.555..

    or about 55.6%

    RonL
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