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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 14% 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 50 cans?

 P(Both defective) = 7/50

 P(Both defective) = 49/2,500

 P(Both defective) = 3/175

 P(Both defective) = 7/25

failurewithasmile  Apr 15, 2017
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P ( 2 defective cans from 50)  =

 

14% defective rate means that 7 cans out of 50 should be defective

 

So.....we have a 7/50 chance of getting a defective can on the first draw   and a 6/49 chance of getting a defective can on the second draw.....so

 

7/50  * 6/49  =

 

7/49 * 6/50  =

 

1/7 *  3/25  =

 

3/175

 

 

 

cool cool cool

CPhill  Apr 15, 2017
edited by CPhill  Apr 15, 2017

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