Question

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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.

Answer 1

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: $p = \frac{5}{{21}}$