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A quality-control analyst at a shipping warehouse has 9 sealed cartons on a shelf, exactly 3 of which contain a damaged item. She selects 2 of the cartons at random, one after another and without replacing the first before drawing the second. What is the probability that at least one of the 2 selected cartons contains a damaged item?
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Correct answer
D
The phrase "at least one" is the cue to find the probability of the opposite outcome and subtract it from 1: P(at least one damaged) = 1 − P(neither damaged). This is faster than adding up the separate cases of exactly one damaged and exactly two damaged.
Method 1 (complement, without replacement). Two facts matter here: 6 of the 9 cartons are undamaged, and the cartons are not replaced between draws. For neither selected carton to be damaged, the first draw must land on one of the 6 undamaged cartons (chance 6/9), and then the second draw must land on one of the 5 undamaged cartons still left among the 8 remaining (chance 5/8). Multiply the two stages, since both must happen: P(neither damaged) = (6/9) × (5/8) = 30/72 = 5/12. So P(at least one damaged) = 1 − 5/12 = 7/12.
Method 2 (estimate and narrow down). Without replacement, P(neither damaged) starts at 6/9, which is two-thirds, and the second factor of 5/8 pulls that down, so P(neither damaged) lands below one-half. That makes the complement, P(at least one damaged), more than one-half. This rules out 1/3 and 5/12, since both sit below one-half. The estimate alone cannot split the remaining three choices, since 5/9, 7/12, and 2/3 all exceed one-half, so finish with the exact complement from Method 1: 1 − 5/12 = 7/12. The other two survivors match known errors instead: 5/9 is what the with-replacement slip produces, and 2/3 is what adding the per-draw chances produces.
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