1986 AJHSME Problems/Problem 24
The 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
Imagine that we run the computer many times. In roughly of all cases, Bob will be assigned to the same group as Al was. Out of these, in roughly cases Carol will be assigned to the same group as her friends. Thus the probability that all three are in the same group is , or .
(The exact value is , which is less than our approximate answer.)
There are ways to choose which group the three kids are in and the chance that all three are in the same group is . Hence or .
One of the statements, that there are students in the school is redundant. Taking that there are students and there are groups, we can easily deduce there are ways to group the students, and there are ways to group them in the same group, so we might think is the answer but as there are 3 groups we do which is .
Solution 4 (easiest)
The information that there are students is irrelevant. The first student has choices to choose from. In order for all students to go in the same lunch group, the second student has choice, and same for the third student.
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