Q1.1.3% of the company's production of a certain component i

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Q1.1.3% of the company's production of a certain component is defective.
(a) what is the probability that the quality inspector will find at 3 defectives when taking a random sample of 74 components?
(b) how large a random sample of components would the inspector need to take to be at least 83% sure of finding at least one defective?

(a) Put P = Probabilty of having 3 defectives in 74 samples
P = C(74,3) x (1.3%)^3 x (100%-1.3%)^(74-3)
= 74! / (74-3)!3! x (1.3%)^3 x (100%-1.3%)^(71)
= (74x73x72) / (3x2x1) x (0.013)^3 x (0.987)^(71)
= 0.0562 or 5.62%
(b) Assume N is the no. of sample tested, 83% = 1 - (100%-1.3%)^N,
0.987^N = 0.17
N log(0.987) = log(0.17)
N = 135.41
Since N is a integral, therefore, N =136.
但要注意你的问题（a) 里面" ... the Quality inspector will find at 3 defectives ... “ 语法不对, 不知道你是要表达"will find 3 defectives" 还是 "will find at least 3 defectives".如果是后者答案就完全不一样.
再问：我少打了个least 应该是后者那应该怎么做你方不方便给我一个联系方式我还有其他2道题目想问

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