SOLUTION: A shipment of 10 items has two defective and eight non defective units.In the inspection of the shipment , a sample of units will be selected and tested. If a defective unit is fou

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Question 1193401: A shipment of 10 items has two defective and eight non defective units.In the inspection of the shipment , a sample of units will be selected and tested. If a defective unit is found , the shipment of 10 units will be rejected. If a sample of three is selected , what is the probability that the shipment will be rejected?
Answer by ikleyn(52781) About Me  (Show Source):
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A shipment of 10 items has two defective and eight non defective units.
In the inspection of the shipment, a sample of units will be selected and tested.
If a defective unit is found, the shipment of 10 units will be rejected.
If a sample of three is selected , what is the probability that the shipment will be rejected?
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The probability that the shipment will be rejected, is the ratio of two numbers.


The denominator is the number of all possible different triples of 10 items,

    C%5B10%5D%5E3 = %2810%2A9%2A8%29%2F%281%2A2%2A3%29 = 120.



The numerator is the number of all triples of 10 item that contain one or two defective items.


The number of triples containing 1 defective item is   C%5B2%5D%5E1.C%5B8%5D%5E2 = 2%2A%28%288%2A7%29%2F2%29 = 8*7 = 56.

The number of triples containing 2 defective items is  C%5B2%5D%5E2.C%5B8%5D%5E1 = 1*8 = 8.


Thus the numerator is  56 + 8 = 64.



Hence, the  ANSWER  to the problem's question is  64%2F120 = 8%2F15.

Solved.