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13 August, 19:48

Question25 warmliquid corp., a manufacturer of water heaters, produces 30 water heaters per week. past records indicate that 10% of total water heaters produced in a week are likely to be defective. suppose a quality check was conducted on a sample of six water heaters. what is the probability that four out of six water heaters will be defective?

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  1. 13 August, 21:22
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    To solve this problem, we use the binomial probability equation.

    P = [n! / (n - r) ! r!] p^r q^ (n - r)

    where,

    n is the total number of samples = 6

    r is the number of defective = 4

    p is the probability it will be defective = 0.10

    q is the probability it wont be = 0.90

    P = [6! / (6 - 4) ! 4!] (0.10) ^4 * (0.90) ^ (6 - 4)

    P = 1.215 x 10^-3 = 0.1215%
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