Difference between revisions of "1986 AJHSME Problems/Problem 24"
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==Solution 2== | ==Solution 2== | ||
| − | One of the statements, | + | One of the statements, that there are <math>600</math> students in the school is redundant. Taking that there are <math>3</math> students and there are <math>3</math> groups, we can easily deduce there are <math>81</math> ways to group the <math>3</math> students, and there are <math>3</math> ways to group them in the same <math>1</math> group, so we might think <math>\frac{3}{54}=\frac{1}{27}</math> is the answer but as there are 3 groups we do <math>\frac{1}{27} (3)=\frac{1}{9}</math> which is <math>\boxed{\text{(B)}}</math>. |
==See Also== | ==See Also== | ||
Revision as of 23:15, 28 November 2020
Contents
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 1
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 2
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 
.
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. 
