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company generally purchases large lots of a certain kind of electronic device.
A method is used that rejects a lot if 2 or more defective units are found
in a random sample of 100 units.
(a) What is the probability of rejecting a lot that is 1% defective?
(b) What is the probability of accepting a lot that is 5% defective?
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In this my post, I will show you the solution for the first problem, ONLY.
In the future, NEVER pack more than ONE problem per post.
This problem can be solved in different ways.
I will show you a way which is among simplest ways, i.e. requires minimum knowledge and minimum calculations.
(a) The probability to get 0 defective units in the random sample of 100 units is
P(0) = = 0.36603 (rounded).
The probability to get 1 defective unit in the random sample of 100 units is
P(1) = = = 0.36973 (rounded)
The probability to get 2 or more defective units is the COMPLEMENT of the sum P(0) + P(1) to 1
P = 1 - P(0) - P(1) = 1 - 0.36603 - 0.36973 = 0.2642 (rounded). ANSWER
It is precisely the probability to reject a lot for given conditions.
(b) The probability to get 0 defective units in the random sample of 100 units is
P(0) = = 0.00592 (rounded).
The probability to get 1 defective unit in the random sample of 100 units is
P(1) = = = 0.03116 (rounded).
The probability to get 2 or more defective units is the COMPLEMENT of the sum P(0) + P(1) to 1
P = 1 - P(0) - P(1) = 1 - 0.00592 - 0.03116 = 0.96292 (rounded). ANSWER
It is precisely the probability to reject a lot for given conditions.
Solved.
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