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2016 AIME I Problems/Problem 2

Revision as of 17:45, 15 October 2021 by Arcticturn (talk | contribs) (→‎Solution 3)

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. We have \[\begin{array}{c||c|c|c|c|c|c|c|c} \boldsymbol{Probability} & \boldsymbol{\frac{1}{21}} & \boldsymbol{\frac{2}{21}} & \boldsymbol{\frac{1}{7}} & \boldsymbol{\frac{4}{21}} & \boldsymbol{\frac{5}{21}} & \boldsymbol{\frac{2}{7}} \\ \hline \hline \hspace{7mm}&\hspace{6.5mm}&\hspace{6.5mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&\hspace{6.75mm}&& \\ [-2.5ex] \hline \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{\frac{1}{21}} &\frac{1}{441} \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{\frac{2}{21}} \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{\frac{1}{7}} \\ \hline &&&&&&&& \\ [-2.25ex] \boldsymbol{\frac{4}{21}} \\ &&&&&&&& \\ [-2.25ex] \boldsymbol{\frac{5}{21}} \\ &&&&&&&& \\ [-2.25ex] \boldsymbol{\frac{2}{7}} \\ \hline \hline &&&&&&&& \\ [-2.25ex] \textbf{Total}&\boldsymbol{3}&\boldsymbol{6}&\boldsymbol{12}&\boldsymbol{24}&\boldsymbol{48}&\boldsymbol{96}&\boldsymbol{192}&\boldsymbol{384} \end{array}\]

I'm still working on it in the mean time PLEASE DO NOT EDIT

~Arcticturn

See also

2016 AIME I (Problems • Answer Key • Resources)
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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