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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

User Abuzze
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1 Answer

4 votes

Answer:

95.69%

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%

User DarrenRhodes
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