1986 AJHSME Problems/Problem 24
Problem
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
Solution 2
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
.
Solution 3
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.
~sakshamsethi
See Also
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 |
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