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Math Help - Probability problem

  1. #1
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    Probability problem

    I was hoping someone could check my work as something seems off to me. Any help/suggestions would be greatly appreciated.

    (a)From a lot of 12 flares, 4 are selected at random. If the lot contains 4 defective flares, what is the probability that all 4 will not work?

    P(4 defective)= C(4,4) / C(12,4)
    = 1/149
    = 0.002 OR 0.2%

    (b)What is the probability that at most 2 will not work?
    P(at most 2 will not work) = P(0, 1 and 2 will not work)

    P (0 will not work) = (C(4,0) * C(8,4)) / C(12,4) = 0.1414...

    P (1 will not work) = (C(4,1) * C(8,3)) / C(12,4) = 0.1131

    P (2 will not work) = (C(4,2) * C(8,2)) / C(12,4) = 0.3394

    Total = 0.5924 or 59.24%


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  2. #2
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    Quote Originally Posted by jake.davis View Post
    I was hoping someone could check my work as something seems off to me. Any help/suggestions would be greatly appreciated.

    (a)From a lot of 12 flares, 4 are selected at random. If the lot contains 4 defective flares, what is the probability that all 4 will not work?

    P(4 defective)= C(4,4) / C(12,4)
    = 1/149
    = 0.002 OR 0.2%

    (b)What is the probability that at most 2 will not work?
    P(at most 2 will not work) = P(0, 1 and 2 will not work)

    P (0 will not work) = (C(4,0) * C(8,4)) / C(12,4) = 0.1414...

    P (1 will not work) = (C(4,1) * C(8,3)) / C(12,4) = 0.1131

    P (2 will not work) = (C(4,2) * C(8,2)) / C(12,4) = 0.3394

    Total = 0.5924 or 59.24%
    Thank you for viewing/responding
    I have not done the calculations.
    But the methods are correct.
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  3. #3
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    Joined
    Sep 2009
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    I took a look at some of your calculations and i saw that for part b.
    P (1 will not work) = (C(4,1) * C(8,3)) / C(12,4) = 0.1131
    when i calculated it i came up with this, i wrote it out so you can see...
    P = 4*56/ 495
    = 224 / 495
    = 0.4525....
    other than that one part the rest of your calculations are correct as far as i've calculated!
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