Difference between revisions of "2016 AIME I Problems/Problem 2"
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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 <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | 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 <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
==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= | + | 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> |
Revision as of 17:48, 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 
