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>
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== See also ==
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{{AIME box|year=2016|n=I|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 16:51, 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 $8+63=071$

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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