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A box of cartridges contains 30 cartridges, of which 6 are defective. If 3 of the cartridges are removed from the box in succession without replacement, what is the probability that all the 3 cartridges are defective?

A box of cartridges contains 30 cartridges, of which 6 are defective. If 3 of the cartridges are removed from the box in succession without replacement, what is the probability that all the 3 cartridges are defective? Correct Answer \(\frac{(6*5*4)}{(30*29*28)}\)

Let A be the event that the first cartridge is defective. Let B be the event that the second cartridge is defective. Let C be the event that the third cartridge is defective. Then probability that all 3 cartridges are defective is P(A ∩ B ∩ C) Hence, P(A ∩ B ∩ C) = P(A) * P(B|A) * P(C | A ∩ B) = (6⁄30) * (5⁄29) * (4⁄28) = (6 * 5 * 4)⁄(30 * 29 * 28).

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