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

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==Problem 2==
 
==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
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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>.
  
 
==Solution==
 
==Solution==
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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
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==Solution 3==
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Since the probability of rolling a <math>1</math> is <math>\frac{1}{21}</math>, the probability of rolling a <math>2</math> is <math>\frac{2}{21}</math> the probability of rolling a <math>3</math> is <math>\frac{3}{21}</math> and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of <math>1</math> and <math>6</math>, <math>2</math> and <math>5</math>, and <math>3</math> and <math>4</math>, and the rest is symmetry and the others are irrelevant.
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We have: <math>2 \cdot (\frac{1}{21} \cdot \frac{2}{7}</math> <math>+</math> <math>\frac{2}{21} \cdot \frac{5}{21}</math> <math>+</math> <math>\frac{1}{7} \cdot \frac{4}{21})</math> <math>=</math> <math>2 \cdot (\frac{2}{147} + \frac{10}{441} + \frac{4}{147})</math> = <math>2 \cdot \frac{4}{63} = \frac{8}{63}</math>. Therefore, the answer is <math>8 + 63</math> = <math>\boxed{071}</math>
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~Arcticturn
  
 
== 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}}

Latest revision as of 13:49, 17 October 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

Solution 3

Since the probability of rolling a $1$ is $\frac{1}{21}$, the probability of rolling a $2$ is $\frac{2}{21}$ the probability of rolling a $3$ is $\frac{3}{21}$ and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of $1$ and $6$, $2$ and $5$, and $3$ and $4$, and the rest is symmetry and the others are irrelevant.

We have: $2 \cdot (\frac{1}{21} \cdot \frac{2}{7}$ $+$ $\frac{2}{21} \cdot \frac{5}{21}$ $+$ $\frac{1}{7} \cdot \frac{4}{21})$ $=$ $2 \cdot (\frac{2}{147} + \frac{10}{441} + \frac{4}{147})$ = $2 \cdot \frac{4}{63} = \frac{8}{63}$. Therefore, the answer is $8 + 63$ = $\boxed{071}$

~Arcticturn

See also

2016 AIME I (Problems • Answer Key • Resources)
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