3 of 6 parts are selected. Here, order is not important
Thus, counting rule that we will use is combinations
Sample space size= 6C3 = 6! / (3!*3!) = 20
P(defective) = 2/6 = 1/3
P(acceptable) = 4/6 = 2/3
Using Binomial theorem,
P(one defective of 3) = 3C1*P(defective)1*P(acceptable)2
= 3 * (1/3) * (2/3)2
= 0.444
Please answer question 4.5 ran- options are available. For each of these three options, another three...
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