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1986 AJHSME Problems/Problem 24

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Problem

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

\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad ...

Solution

Imagine that we run the computer many times. In roughly 1/3 of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly 1/3 cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is \frac{1}{3}\cdot \frac{1}{3} = \frac{1}{9}, or \boxed{\text{B}}.

(The exact value is \frac{199}{599} \cdot \frac{198}{598}, which is \sim 0.11\% less than our approximate answer.)

See Also

1986 AJHSME (ProblemsResources)
Preceded by
Problem 23
Followed by
Problem 25
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