Question 1199388
.
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.



<pre>
(a)  The probability to get 0 defective units in the random sample of 100 units is  

                     P(0) = {{{(1-0.01)^100}}} = 0.36603  (rounded).


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

                     P(1) = {{{C[100]^1*0.01(1-0.01)^99}}} = {{{100*0.01(1-0.01)^99}}} = 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).   <U>ANSWER</U>


     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) = {{{(1-0.05)^100}}} = 0.00592  (rounded).


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

                     P(1) = {{{C[100]^1*0.01(1-0.05)^99}}} = {{{100*0.05(1-0.05)^99}}} = 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).   <U>ANSWER</U>


     It is precisely the probability to reject a lot for given conditions.
</pre>

Solved.


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