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