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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,
= = 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 . = = 8*7 = 56.
The number of triples containing 2 defective items is . = 1*8 = 8.
Thus the numerator is 56 + 8 = 64.
Hence, the ANSWER to the problem's question is = .
Solved.