1986 AJHSME Problems/Problem 24

Revision as of 14:05, 17 April 2020 by Ryjs (talk | contribs) (Solution 2)


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.)

Solution 2

There are $\binom{3}{1}$ ways to choose which group the three kids are in and the chance that all three are in the same group is $\frac{1}{27}$. Hence $\frac{1}{9}$ or $\boxed {B}$.

Solution 3

One of the statements, that there are $600$ students in the school is redundant. Taking that there are $3$ students and there are $3$ groups, we can easily deduce there are $81$ ways to group the $3$ students, and there are $3$ ways to group them in the same $1$ group, so we might think $\frac{3}{54}=\frac{1}{27}$ is the answer but as there are 3 groups we do $\frac{1}{27} (3)=\frac{1}{9}$ which is $\boxed{\text{(B)}}$.

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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