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A shipment of 9 printers contains 3 that are defective. Find the probability that a sample of size 3, drawn from the 9, will not contain a defective printer.
3 printers out of 10 can be chosen "n = C_9^3" ways - total number of outcomes.
The number of favorable outcomes is "m = C_{9 - 3}^3 = C_6^3".
Then the wanted probability is
"p = \\frac{m}{n} = \\frac{{C_6^3}}{{C_9^3}} = \\frac{{6!}}{{3!3!}} \\cdot \\frac{{3!6!}}{{9!}} = \\frac{{4 \\cdot 5 \\cdot 6}}{{7 \\cdot 8 \\cdot 9}} = \\frac{5}{{7 \\cdot 3}} = \\frac{5}{{21}}"
Answer: "p = \\frac{5}{{21}}"
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