Difference between revisions of "1983 AIME Problems/Problem 7"
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The total probability is <math>\frac{420}{552}=\frac{35}{46}</math>, so our answer is <math>1-\frac{35}{46}=\frac{11}{46}</math>. | The total probability is <math>\frac{420}{552}=\frac{35}{46}</math>, so our answer is <math>1-\frac{35}{46}=\frac{11}{46}</math>. | ||
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== See also == | == See also == | ||
| + | {{AIME box|year=1983|num-b=6|num-a=8}} | ||
* [[AIME Problems and Solutions]] | * [[AIME Problems and Solutions]] | ||
* [[American Invitational Mathematics Examination]] | * [[American Invitational Mathematics Examination]] | ||
Revision as of 17:38, 21 March 2007
Problem
Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let 
be the brobability that at least two of the three had been sitting next to each other. If is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?
Solution
We can also find the probability that none of them are sitting together, and subtract from 1. There are 2 cases:
- They are all separated by at least 2 seats.
- One of them is separated from one of the others by 1 seat.
In case (1), the probability is 
.
But what if they're only separated by 1 seat? In that case, the probability is 
.
The total probability is 
, so our answer is 
.
See also
| 1983 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
