Difference between revisions of "1986 AJHSME Problems/Problem 24"
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(The exact value is <math>\frac{199}{599} \cdot \frac{198}{598}</math>, which is <math>\sim 0.11\%</math> less than our approximate answer.) | (The exact value is <math>\frac{199}{599} \cdot \frac{198}{598}</math>, which is <math>\sim 0.11\%</math> less than our approximate answer.) | ||
==Solution 2== | ==Solution 2== | ||
− | There are <math> | + | There are <math>\binom{3}{1}</math> ways to choose which group the three kids are in and the chance that all three are in the same group is <math>\frac{1}{27}</math>. Hence <math>\frac{1}{9}</math> or <math>\boxed {B}</math>. |
==Solution 3== | ==Solution 3== |
Revision as of 14:05, 17 April 2020
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 .
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.