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Yesterday, 14:47

A lot of size n = 30 contains three nonconforming units. what is the probability that a sample of five units selected at random contains exactly one nonconforming unit? what is the probability that it contains one or more nonconformances?

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  1. Yesterday, 16:48
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    Exactly 1 nonconforming unit = 0.369458128

    1 or more nonconforming units = 0.433497537

    Due to formatting issues, I'll use the notion C (n, x) for N choose X. Which would be n! / (x! (n-x) !).

    Now for the case of exactly 1, the probability will be:

    P = C (3,1) * C (30-3,4) / C (30,5)

    To explain it, you choose exactly 1 nonconforming item out of the 3 possible, then fill the sample with 4 more conforming items to complete the sample size of 5. Finally, you divide by the number of different ways you can select 5 items out of the entire group of 30. So doing the math, you get P = C (3,1) * C (30-3,4) / C (30,5) = 3*17550/142506 = 0.369458128 = 36.9458128%

    For the case of 1 or more nonconforming units you do the sum of x ranging from 1 to 3 with the formula

    C (3, x) * C (30-3,5-x) / C (30,5)

    or you can set x to 0 and evaluate

    1 - C (3,0) * C (30-3,5-0) / C (30,5)

    Let's do it both ways.

    C (3,1) * C (30-3,5-1) / C (30,5) +

    C (3,2) * C (30-3,5-2) / C (30,5) +

    C (3,3) * C (30-3,5-3) / C (30,5)

    = 3*17550/142506 + 3*2925/142506 + 1*351/142506

    = 0.369458128 + 0.061576355 + 0.002463054

    = 0.433497537

    And doing it the "simple" way

    1 - C (3,0) * C (30-3,5-0) / C (30,5)

    = 1 - 1*80730/142506 = 1 - 0.566502463 = 0.433497537
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