Problem 24 · 1986 AJHSME
Stretch
Counting & Probability
fix-one-condition-rest
The 600 students at King Middle School are divided into three groups of equal size for lunch. Each group has lunch at a different time. A computer randomly assigns each student to one of three lunch groups. The probability that three friends, Al, Bob, and Carol, will be assigned to the same lunch group is approximately
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Answer: B — 1⁄9.
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Hint 1 of 3
Don't worry about *which* of the three groups they all end up in — just let Al go wherever he goes and treat his group as 'the target.' Now the only question is whether the other two follow him there.
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Hint 2 of 3
Each remaining friend independently lands in Al's group with chance about 1⁄3. The word 'same group' means *both* of them must match Al, so combine the two chances.
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Hint 3 of 3
Independent 'and' events multiply their probabilities.
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Approach: anchor on Al, then require the others to match
- Anchor the problem on Al: wherever he lands becomes 'the target group,' which costs no probability — he's somewhere for sure. This sidesteps having to sum over all three groups.
- Now Bob must land in Al's group (chance ≈ 1⁄3, since the three groups are equal size and the 600 students make it essentially even) and Carol must too (another ≈ 1⁄3), independently.
- 'Bob matches AND Carol matches' multiplies: 1⁄3 × 1⁄3 = 1⁄9.
- Why anchoring helps: fixing one object as the reference removes a layer of casework — you no longer care which group, only that the rest agree with it. (The '600 students, equal groups' detail is what makes each chance a clean 1⁄3.)
Another way — count equally likely group-triples:
- List the group each friend gets as a triple; there are 3 × 3 × 3 = 27 equally likely assignments. 'All same' happens in 3 of them (all-group-1, all-group-2, all-group-3), so the probability is 3⁄27 = 1⁄9.
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