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 <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>\frac{56}{441}=\frac{8}{63}</math>.  The answer is <math>8+63=\boxed{071}</math>
 
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>\frac{56}{441}=\frac{8}{63}</math>.  The answer is <math>8+63=\boxed{071}</math>
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See also [[2006 AMC 12B Problems/Problem 17]]
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2016|n=I|num-b=1|num-a=3}}
 
{{AIME box|year=2016|n=I|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:09, 9 June 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 $\frac{56}{441}=\frac{8}{63}$. The answer is $8+63=\boxed{071}$

See also 2006 AMC 12B Problems/Problem 17

See also

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions

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