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QUESTION
A box contains 10 computer disks. If it is known that exactly 1 of
the disks is faulty what is the probability that none of the first
three disks taken from the box, without replacement, are faulty?
ANSWER
$p($first 3 not faulty$)\\=p($first not faulty and second not
faulty and third not
faulty$\ds)\\=\frac{9}{10}\times\frac{8}{9}\times\frac{7}{8}=\frac{7}{10}$
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