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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 count the opposite outcome and subtract: P(at least one damaged) = 1 - P(neither damaged). This is faster than adding the separate cases of exactly one and exactly two damaged.
Method 1 (complement, without replacement). Underline two facts in the stem: 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 be one of the 6 undamaged (chance 6/9) and then the second draw must be one of the 5 undamaged still left out of the 8 remaining (chance 5/8). Multiply, 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 eliminate). Without replacement, P(neither damaged) is a bit under one-half (you start at 6/9, which is two-thirds, then the second factor 5/8 pulls it down), so the complement is a bit over one-half. That immediately rules out 1/3 and 5/12 (both below one-half) and points to a value just past 1/2. Of the remaining choices, 7/12 (just over one-half) fits, while 5/9 and 2/3 sit higher and correspond to the with-replacement and add-the-chances errors.
Why the distractors are tempting: 1/3 stops at the chance one carton is damaged and forgets that two are drawn. 5/12 does the without-replacement work correctly but forgets the final step of subtracting from 1, the most common slip on "at least one" problems. 5/9 forgets that the first carton is kept out, using 6/9 again for the second draw as if it were replaced. 2/3 adds the two single-draw damaged chances, double-counting the case where both cartons are damaged, which over-states the probability.
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