Difference between revisions of "2016 AIME I Problems/Problem 2"
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==Solution== | ==Solution== | ||
It is easier to think of the dice as 21 sided dice with 6 sixes, 5 fives, etc. Then there are 21^2=441 possible roles. There are 2*(1*6+2*5+3*4)=56 roles that will result in a seven. The odds are therefore <math>56/441=8/63</math>. The answer is <math>8+63=071</math> | It is easier to think of the dice as 21 sided dice with 6 sixes, 5 fives, etc. Then there are 21^2=441 possible roles. There are 2*(1*6+2*5+3*4)=56 roles that will result in a seven. The odds are therefore <math>56/441=8/63</math>. The answer is <math>8+63=071</math> | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2016|n=I|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:51, 4 March 2016
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 
and 
are relatively prime positive integers. Find 
.
Solution
It is easier to think of the dice as 21 sided dice with 6 sixes, 5 fives, etc. Then there are 21^2=441 possible roles. There are 2*(1*6+2*5+3*4)=56 roles that will result in a seven. The odds are therefore 
. The answer is 
See also
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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. 
