# 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.)

There are $3C1$ 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 $\boxed {B}$.