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# A manufacturer produces soda cans...

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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

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

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