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anonymous

  • one year ago

pleeeeeeeeeeeeaaaaaaaase helllllp me.......... a certain machine make matches. one match in 10000 on average is defective. using a suitable approximation, find the probability that a random sample of 45000 matches will include 2, 3 or 4 defective matches

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  1. anonymous
    • one year ago
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    @nincompoop

  2. anonymous
    • one year ago
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    @IrishBoy123

  3. anonymous
    • one year ago
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    @wio

  4. anonymous
    • one year ago
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    yesss.. actually am learning poisson distribution

  5. anonymous
    • one year ago
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    and ?

  6. alekos
    • one year ago
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    P(2 defects) = P(1 defect) x P(1 defect) P(3 defects) = P(2 defects) x P(1 defect) P(4 defects) = P(3 defects) x P( 1 defect)

  7. anonymous
    • one year ago
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    how would you calculate them ?

  8. alekos
    • one year ago
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    P(2 or 3 or 4 defects) = P(2 defects) + P(3 defects) + P(4 defects)

  9. anonymous
    • one year ago
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    help me with the numbers

  10. alekos
    • one year ago
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    well you already know that P(1 defect) = 1/10000 so from there it is just simple arithmetic

  11. anonymous
    • one year ago
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    yess

  12. alekos
    • one year ago
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    use your calculator in exponential or scientific mode

  13. anonymous
    • one year ago
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    are you sure its simple arithmetic ? hve u calculated ?

  14. alekos
    • one year ago
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    yes, I sure have

  15. anonymous
    • one year ago
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    what ans uve got ?

  16. alekos
    • one year ago
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    Approximately \[\frac{ 1 }{ 10^{8} }\]

  17. anonymous
    • one year ago
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    its wrong.. the answer should be 0.471 its not simple arithmetic

  18. anonymous
    • one year ago
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    aniway thanks

  19. alekos
    • one year ago
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    can you send me a photo of the actual question?

  20. anonymous
    • one year ago
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    it is the actual question

  21. alekos
    • one year ago
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    That figure is not possible based on the information you have given

  22. IrishBoy123
    • one year ago
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    i can help you but not whilst others are as that is counter productive for all concerned tag me later

  23. anonymous
    • one year ago
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    do you know poisson distribution ?

  24. alekos
    • one year ago
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    go for it IrishBoy

  25. IrishBoy123
    • one year ago
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    yes and the answer is 0.471004095.......

  26. IrishBoy123
    • one year ago
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    thx @alekos :p

  27. IrishBoy123
    • one year ago
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    start with the formula and work out \( \lambda \). have you done that? expected outcome given failure rate of 1/10000

  28. anonymous
    • one year ago
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    yes

  29. IrishBoy123
    • one year ago
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    then just plug in as per @alekos suggestion ie P(2 or 3 or 4 defects) = P(2 defects) + P(3 defects) + P(4 defects) but what is your \( \lambda\)? that's all that's missing

  30. IrishBoy123
    • one year ago
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    \( \lambda\) is how many you would expect to be defective in a lot of 45,000 given failure rate 1/10,000 then \( P(X=2) = \frac{e^{- \lambda } \lambda^2}{2!}\) do same for 3 and 4 and add and you're done

  31. anonymous
    • one year ago
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    thank you

  32. IrishBoy123
    • one year ago
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    i have to go but here are the numbers you should get 2 0.11247859 3 0.168717885 4 0.189807621 \(\Sigma\) 0.471004095

  33. anonymous
    • one year ago
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    yeah.. got all

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