Suppose that the probability of a DVD player that was manufactured in a certain factory being defective is 2%. What is the probability that 3 DVD players are defective in a shipment of 13 DVD players from this factory?

Use the binomial probability formula for exactly 3 defective players: choose 3 of the 13, then multiply by the probability those 3 are defective and the other 10 are not. This gives C(13,3)(0.02)^3(0.98)^10 ≈ 0.00187, which rounds to 0.0019. So the correct choice is 0.0019.

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