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A manufacturer produces soda cans and a quality control worker randomly selects two cans from the assembly line for testing. Past statistics show that 10% of the cans are defective. What is the probability that the two selected cans are defective if the quality control worker selects the two cans from a batch of 40 cans?

 

P(Both defective) = 1/130

P(Both defective) =23/130

P(Both defective) =1/100

P(Both defective) = 1/5

hayleeirene  Apr 28, 2018
 #1
avatar+87333 
+1

 

Considering each can to be unique, the number of diiferent  sets of cans  we can form on two successive draws  is C(40,2)  =  780

 

And since  10% of the cans are defective......this means that we have 4 defective cans......and we want to choose any 2 of them ....so   .....  C(4,2)  = 6   possible sets consisting of only defective cans

 

So....the probability that  any set we choose  will contain  2 defective cans  is :

 

Sets containing defective cans  / Total possible sets  =  

 

6 / 780   =   

 

1/ 130

 

 

cool cool cool

CPhill  Apr 28, 2018
edited by CPhill  Apr 28, 2018

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