Difference between revisions of "2016 AIME I Problems/Problem 2"
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==Problem 2== | ==Problem 2== | ||
| − | Two dice appear to be normal dice with their faces numbered from <math>1</math> to <math>6</math>, but each die is weighted so that the probability of rolling the number <math>k</math> is directly proportional to <math>k</math>. The probability of rolling a <math>7</math> with this pair of dice is <math>\frac{m}{n}</math>, where | + | Two dice appear to be normal dice with their faces numbered from <math>1</math> to <math>6</math>, but each die is weighted so that the probability of rolling the number <math>k</math> is directly proportional to <math>k</math>. The probability of rolling a <math>7</math> with this pair of dice is <math>\frac{m}{n}</math>, where |
| + | |||
==Solution== | ==Solution== | ||
It is easier to think of the dice as <math>21</math> sided dice with <math>6</math> sixes, <math>5</math> fives, etc. Then there are <math>21^2=441</math> possible rolls. There are <math>2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56</math> rolls that will result in a seven. The odds are therefore <math>\frac{56}{441}=\frac{8}{63}</math>. The answer is <math>8+63=\boxed{071}</math> | It is easier to think of the dice as <math>21</math> sided dice with <math>6</math> sixes, <math>5</math> fives, etc. Then there are <math>21^2=441</math> possible rolls. There are <math>2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56</math> rolls that will result in a seven. The odds are therefore <math>\frac{56}{441}=\frac{8}{63}</math>. The answer is <math>8+63=\boxed{071}</math> | ||
Revision as of 06:36, 3 June 2018
Problem 2
Two dice appear to be normal dice with their faces numbered from 
to 
, but each die is weighted so that the probability of rolling the number 
is directly proportional to . The probability of rolling a 
with this pair of dice is 
, where
Solution
It is easier to think of the dice as 
sided dice with sixes, 
fives, etc. Then there are 
possible rolls. There are 
rolls that will result in a seven. The odds are therefore 
. The answer is 
See also 2006 AMC 12B Problems/Problem 17
See also
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
