Difference between revisions of "2006 AIME II Problems/Problem 5"
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0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\ | 0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\ | ||
A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*}</cmath> | A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*}</cmath> | ||
− | We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>. Therefore, the probability is <math>\frac{5}{24}</math> and the answer is <math>5+24=\boxed{ | + | We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>. Therefore, the probability is <math>\frac{5}{24}</math> and the answer is <math>5+24=\boxed{29}</math>. |
Note also that the initial assumption that face <math>F</math> 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 <math>A(6)</math> as <math>p</math>, for example, and replaced the others with variables too, but the notation would have been harder to follow. | Note also that the initial assumption that face <math>F</math> 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 <math>A(6)</math> as <math>p</math>, for example, and replaced the others with variables too, but the notation would have been harder to follow. | ||
== Solution 2 == | == Solution 2 == | ||
− | + | We have that the cube probabilities to land on its faces are <math>\frac{1}{6}</math>, <math>\frac{1}{6}</math>, <math>\frac{1}{6}</math>, <math>\frac{1}{6}</math> ,<math>\frac{1}{6}+x</math> ,<math>\frac{1}{6}-x </math> | |
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: | 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: | ||
− | <cmath>4 | + | <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> | ||
− | so the probability F which is greater than <math>\frac{1}{6}</math> is equal <math>\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{ | + | so the probability F which is greater than <math>\frac{1}{6}</math> is equal <math>\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{29}</math> |
+ | == Solution 3 (Alcumus) == | ||
+ | Let <math>p(a,b)</math> denote the probability of obtaining <math>a</math> on the first die and <math>b</math> on the second. Then the probability of obtaining a sum of 7 is<cmath>p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).</cmath>Let the probability of obtaining face <math>F</math> be <math>(1/6)+x</math>. Then the probability of obtaining the face opposite face <math>F</math> is <math>(1/6)-x</math>. Therefore | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | {{47}\over{288}}&= | ||
+ | 4\left({1\over6}\right)^2+2\left({1\over6}+x\right) | ||
+ | \left({1\over6}-x\right)\cr&= | ||
+ | {4\over36}+2\left({1\over36}-x^2\right)\cr&= | ||
+ | {1\over6}-2x^2. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Then <math>2x^2=1/288</math>, and so <math>x=1/24</math>. The probability of obtaining face <math>F</math> is therefore <math>(1/6)+(1/24)=5/24</math>, and <math>m+n=\boxed{29}</math>. | ||
== See also == | == See also == |
Latest revision as of 11:10, 5 November 2023
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
Solution 3 (Alcumus)
Let denote the probability of obtaining on the first die and on the second. Then the probability of obtaining a sum of 7 isLet the probability of obtaining face be . Then the probability of obtaining the face opposite face is . Therefore
Then , and so . The probability of obtaining face is therefore , and .
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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