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A lot contains 12 items and 4 are defective. If 3 items are drawn at random from the lot, what is the probability none of them are defective?

12 - 4 = 8

8/12 * 7/11 * 6/10 = .6667 * .6663 * .6000 =


Not sure if I am on the right track - any suggestions?

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$\frac 8{12} \ne 0.6667$ and the closest four digit expansion of $\frac 7{11}$ is $0.6364$ (typo plus misrounding). I would leave things as fractions and report the result as $\frac {14}{55}$ or say it was $\approx 0.2545$ – Ross Millikan Mar 8 '13 at 14:20

$12$ total, $4$ defective, $8$ non-defective

No. of ways of selecting $3$ non-defective balls out of $8={8\choose 3}$

Total no. of ways of selecting $3$ balls = ${12\choose 3}$

Thus, Required probability = $\frac{8\choose 3}{12\choose 3}$ and when you simplify it , it would be same as your answer(which is correct).

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