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15 January, 05:52

Suppose parts are of two varieties: good (with probability 90/92) and slightly defective (with probability 2/92). Parts are produced one after the other. What is the probability that at least 3 parts must be produced until there is a slightly defective part produced

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  1. 15 January, 07:24
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    95.69%

    Step-by-step explanation:

    We have X is the number of parts produced up to (and including) the first slightly defective part. So, X is Geometric (2/92), which would be the following:

    P (X = > 3) = Summation i = 3, up to infinity of {[ (90/92) ^ (i-1) ] * (2/92) }

    We replace and solve and we are left with:

    P (X = > 3) = (2/92) * (90/92) ^ (3-1) * 1 / (1 - 90/92)

    P (X = > 3) = 0.9569

    Which means that the probability that at least 3 parts must be produced until there is a slightly defective part produced is 95.69%
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