Homework Answers
2) p(1 defective 1 working) = 
So, 4/7 is the probability for the first time selection of 1 defective and 1 working.
It will remain same for other two turns also.
So, P(three times 1 defective, 1 working) = 
So, 64/343 is the probability that one defective and one working are chosen each time.
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