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A box contain 12 cell phones of which 4 are defective.all phones look alike.3 phones are drawn randomly what is the probability.1 all the 3 bulbs are defective.2 at least 2 of the bulbs chosen are defective.3 at most 2 of the bulbs chosen are defective.hint use combination analysis

  • math -

    prob(defective) = 4/12 = 1/3
    prob(not defective) = 2/3

    1. all 3 defective ---- prob = (1/3)^3 = 1/27
    2. at least 2 defective
    ----> 2 defective or 3 defective
    prob = C(3,2)(1/3)^2 (2/3) + (1/3)^3
    = 3(2/27) + 1/27 = 7/27

    3. at most 2 ---> 0 defective, 1 defect, or 2 defective
    or exclude all 3 defectiv
    = 1 - 1/27 = 26/27

    BTW,
    Where do the bulbs enter the picture when talking about cell phones

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