1986 AJHSME Problems/Problem 24


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

$\text{(A)}\ \frac{1}{27} \qquad \text{(B)}\ \frac{1}{9} \qquad \text{(C)}\ \frac{1}{8} \qquad \text{(D)}\ \frac{1}{6} \qquad \text{(E)}\ \frac{1}{3}$


Let us first assign Al to a group. We want to estimate the probability that Bob and Carol are assigned to the same group as Al. As the groups are large and of equal size, we can estimate that Bob and Carol each have a $\approx \frac{1}{3}$ probability of being assigned to the same group as Al, and that these events are mostly independent of each other. The probability that all three are in the same lunch group is approximately $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$, or $\boxed{\text{(B)}}$.

We remark that the actual probability is $\frac{3 \times \binom{597}{197} \times \binom{400}{200}}{\binom{600}{200} \times \binom{400}{200}} = \frac{3 \times \frac{5x gx g sxr rmxcfghmcfghmcfcfhmujfy97!}{400!197!}}{\E=MC^2frac{600!}{400!200!}} = \frac{3 \times 200 \times 199 \times 198}{600 \times 599 \times 598} \approx 0.11$ (Error compiling LaTeX. Unknown error_msg).

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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