Difference between revisions of "2016 AIME I Problems/Problem 2"

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See also [[2006 AMC 12B Problems/Problem 17]]
 
See also [[2006 AMC 12B Problems/Problem 17]]
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==Solution 2==
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Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling <math>n</math> is <math>\frac{n}{21}</math> because <math>21=\frac{6 \cdot 7}{2}</math>
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Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice.
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The probability that 7 is rolled is now <math>2(\frac{1}{21} \cdot \frac{6}{21}+\frac{2}{21} \cdot \frac{5}{21} + \frac{3}{21} \cdot \frac{4}{21})</math> which is equal to <math>\frac{56}{441}=\frac{8}{63}</math>.  Therefore the answer is <math>8+63=\boxed{071}</math>
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~PEKKA
  
 
== 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 15:25, 12 August 2021

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

Solution 2

Since the probability of rolling any number is 1, and the problem tells us the dice are unfair, we can assign probabilities to the individual faces. The probability of rolling $n$ is $\frac{n}{21}$ because $21=\frac{6 \cdot 7}{2}$ Next, we notice that 7 can be rolled by getting individual results of 1 and 6, 2 and 5, or 3 and 4 on the separate dice. The probability that 7 is rolled is now $2(\frac{1}{21} \cdot \frac{6}{21}+\frac{2}{21} \cdot \frac{5}{21} + \frac{3}{21} \cdot \frac{4}{21})$ which is equal to $\frac{56}{441}=\frac{8}{63}$. Therefore the answer is $8+63=\boxed{071}$ ~PEKKA

See also

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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