Difference between revisions of "1986 AJHSME Problems/Problem 24"

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==See Also==
==See Also==
[[1986 AJHSME Problems]]
{{AJHSME box|year=1986|num-b=23|num-a=25}}
[[Category:Probability Problems]]

Revision as of 19:16, 23 May 2009


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

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


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 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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All AJHSME/AMC 8 Problems and Solutions
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