Difference between revisions of "2016 AIME I Problems/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 <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | 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== | ||
− | 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{ | + | 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{71}</math> |
See also [[2006 AMC 12B Problems/Problem 17]] | See also [[2006 AMC 12B Problems/Problem 17]] |
Revision as of 21:12, 4 December 2016
Problem 2
Two dice appear to be normal dice with their faces numbered from to , but each die is weighted so that the probability of rolling the number is directly proportional to . The probability of rolling a with this pair of dice is , where and are relatively prime positive integers. Find .
Solution
It is easier to think of the dice as sided dice with sixes, fives, etc. Then there are possible rolls. There are rolls that will result in a seven. The odds are therefore . The answer is
See also 2006 AMC 12B Problems/Problem 17
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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