A box contains 10 bulbs, 4 of which are defective. If 3 of the bulbs are drawn randomly without…
Probability=(4/10)(3/9)(2/8)=24/720=1/30, equivalently C(4,3)/C(10,3)=4/120. 4/10 is just the first-draw defect probability; 3/10 and 3/4 do not count the sequential…
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A box contains 10 bulbs, 4 of which are defective. If 3 of the bulbs are drawn randomly without replacement, what is the probability that all three bulbs are defective?
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Correct answer A
Probability=(4/10)(3/9)(2/8)=24/720=1/30, equivalently C(4,3)/C(10,3)=4/120. 4/10 is just the first-draw defect probability; 3/10 and 3/4 do not count the sequential without-replacement event.
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UGC NET Economics