1986 AJHSME Problems/Problem 24

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

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

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.)

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 $\boxed{\text{(B)}}$ .

Solution 4

1986 AJHSME (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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