Difference between revisions of "2006 AIME II Problems/Problem 5"
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<cmath>4 \cdot \left(\frac{1}{6} \right)^2+2 \left(\frac{1}{6}+x \right) \left(\frac{1}{6}-x \right)=\frac{47}{288}</cmath> | <cmath>4 \cdot \left(\frac{1}{6} \right)^2+2 \left(\frac{1}{6}+x \right) \left(\frac{1}{6}-x \right)=\frac{47}{288}</cmath> | ||
multiplying by 288 we get: | multiplying by 288 we get: | ||
− | <cmath>32+16(6x | + | <cmath>32+16(1-6x)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15</cmath> |
dividing by 16 and rearranging we get: | dividing by 16 and rearranging we get: | ||
<cmath>\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}</cmath> | <cmath>\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}</cmath> |
Latest revision as of 12:49, 7 March 2021
Contents
Problem
When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face is greater than , the probability of obtaining the face opposite is less than , the probability of obtaining any one of the other four faces is , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is . Given that the probability of obtaining face is where and are relatively prime positive integers, find
Solution 1
Without loss of generality, assume that face has a 6, so the opposite face has a 1. Let be the probability of rolling a number on one die and let be the probability of rolling a number on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of , totaling . Subtracting all these probabilities from leaves chance of getting a 1 on die and a 6 on die or a 6 on die and a 1 on die :
Since the two dice are identical, and so
Also, we know that and that the total probability must be , so:
Combining the equations:
We know that , so it can't be . Therefore, the probability is and the answer is .
Note also that the initial assumption that face was the face labelled 6 is unnecessary -- we would have carried out exactly the same steps and found exactly the same probability no matter which face it was. We could have labelled as , for example, and replaced the others with variables too, but the notation would have been harder to follow.
Solution 2
We have that the cube probabilities to land on its faces are , , , , , we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: multiplying by 288 we get: dividing by 16 and rearranging we get: so the probability F which is greater than is equal
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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