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=071</math>
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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>56/441=8/63</math>.  The answer is <math>8+63=\boxed{071}</math>
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:53, 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 rolls. There are $2\cdot(1\cdot 6+2\cdot 5+3\cdot 4)=56$ rolls that will result in a seven. The odds are therefore $56/441=8/63$. The answer is $8+63=\boxed{071}$

See also

2016 AIME I (ProblemsAnswer KeyResources)
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

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