2016 AIME I Problems/Problem 2
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
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 is because 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 which is equal to . Therefore the answer is ~PEKKA
Solution 3
Since the probability of rolling a is , the probability of rolling a is the probability of rolling a is and so on, we can make a chart of probabilities and add them together. Note that we only need the probabilities of and , and , and and , and the rest is symmetry and the others are irrelevant.
We have: = . Therefore, the answer is =
~Arcticturn
See also
2016 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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