SOLUTION: (1) 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 o

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Question 1199388: (1) 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?

(2) The acceptance scheme for purchasing lots containing a large number of batteries is to test no more than 75 randomly selected batteries
and to reject a lot if a single battery fails. Suppose the probability of a failure is 0.001.
(a) What is the probability that a lot is accepted?
(b) What is the probability that a lot is rejected on the 20th test?
(c) What is the probability that it is rejected in 10 or fewer trials?
Answer by ikleyn(52847) About Me  (Show Source):
You can put this solution on YOUR website!
.
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) = %281-0.01%29%5E100 = 0.36603  (rounded).


     The probability to get 1 defective unit  in the random sample of 100 units is  

                     P(1) = C%5B100%5D%5E1%2A0.01%281-0.01%29%5E99 = 100%2A0.01%281-0.01%29%5E99 = 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) = %281-0.05%29%5E100 = 0.00592  (rounded).


     The probability to get 1 defective unit  in the random sample of 100 units is  

                     P(1) = C%5B100%5D%5E1%2A0.01%281-0.05%29%5E99 = 100%2A0.05%281-0.05%29%5E99 = 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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