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Binomial Distribution Probability

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data
    A quality control engineer tests the quality of produced computers. Suppose that 5% of computers have defects, and defects occur independently of each other.

    A- What is the expected number of defective computers in a shipment of twenty?
    B- Find the probability of exactly 3 defective computers in a shipment of twenty.
    C- Find the probability that the engineer has to test at least 5 computers in order to find a defective one.


    2. Relevant equations



    3. The attempt at a solution

    A-1
    B- .061 which were correct

    For C I got .00257 which was wrong. I just don't understand how I'm supposed to find at least 5 when I don't know the total number since for finding .00257 I used n=20 which I guess I wasn't supposed to since I got it wrong
     
  2. jcsd
  3. Feb 20, 2012 #2

    tiny-tim

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    hi lina29! :smile:
    with questions like this, it often helps to write out the opposite, and find the probability of that

    so how would you write (in english) the opposite of "the engineer has to test at least 5 computers in order to find a defective one" ? :wink:
     
  4. Feb 20, 2012 #3
    the engineer has to test at most 4 computers in order to find a defective one
     
  5. Feb 20, 2012 #4

    tiny-tim

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    oops!

    oops! :redface:

    yes, that's right, i asked the wrong question :rolleyes:

    i should have asked you just to rewrite the question in terms of successes, rather than failures :smile:
     
  6. Feb 20, 2012 #5

    Ray Vickson

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    Please show your work. Where, exactly, does the number 0.00257 come from? Without this information it is impossible to help you; we cannot tell if you used the correct formula but made an arithmetical error, or what.

    RGV
     
  7. Feb 20, 2012 #6
    using my calculator I did 1-binomcdf(20,.05,4)
     
  8. Feb 20, 2012 #7
    I don't know how I would write it in terms of successes rather than failures
     
  9. Feb 20, 2012 #8

    tiny-tim

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    so how many good ones will there be? :wink:
     
  10. Feb 20, 2012 #9
    15? assuming that C also is also doing a shipment of 20 computers. So would I find exactly 15 successes?
     
  11. Feb 20, 2012 #10

    tiny-tim

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    if he gets to the 5th computer, what happened to the first 4 ?
     
  12. Feb 20, 2012 #11
    they were nondefective
     
  13. Feb 20, 2012 #12

    tiny-tim

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    yes! :smile:

    so probability that the engineer has to test at least 5 computers in order to find a defective one = probability that the first 4 are not defective :wink:

    which is … ? :smile:
     
  14. Feb 20, 2012 #13
    (.95)^4=.81451
     
  15. Feb 20, 2012 #14

    tiny-tim

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    ok …

    so isn't that the answer?​
     
  16. Feb 20, 2012 #15
    lol I was just making sure it was the right answer before I put it in and it is :) Thanks!
     
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