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=71</math>
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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 16:48, 4 March 2016

Problem 2

Two dice appear to be normal dice with their faces numbered from $1$ to $6$, but each die is weighted so that the probability of rolling the number $k$ is directly proportional to $k$. The probability of rolling a $7$ with this pair of dice is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

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 $56/441=8/63$. The answer is $8+63=071$

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