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There are 200 items in a box and 5 % of them are defective. In how many ways can 3 items can be selected from the box so that all the items selected are defective?

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5 % of items are defective in a box of 200 items.

∴ No. of defective items in the box = 200 × 5 / 100 =10 items.

3 items are selected from the box. Such that all three items are defective.

∴ Total combinations = 10C2

\(\frac{10×9×8}{3×2×1}\)

= 720 / 6 = 120

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