Difference between revisions of "2006 AIME I Problems/Problem 5"
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== Solution == | == Solution == | ||
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− | <math> a\ | + | For now, assume that face <math>F</math> has a 6, so the opposite face has a 1. Let <math>A(n)</math> be the probability of rolling a number <math>n</math> on one die and let <math>B(n)</math> be the probability of rolling a number <math>n</math> on the other die. 7 can be obtained by rolling a <math>A(n)=2</math> and <math>B(n)=5</math>, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of <math>\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}</math>, totaling <math>4 \cdot \frac{1}{36} = \frac{1}{9}</math>. Subtracting all these probabilities from <math>\frac{47}{288}</math> leaves <math>\frac{15}{288}=\frac{5}{96}</math> chance of getting a 1 on die <math>A</math> and a 6 on die <math>B</math> or a 6 on die <math>A</math> and a 1 on die <math>B</math>: |
− | + | <math>A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}</math> | |
− | <math> | + | Since the two dice are identical, <math>B(1)=A(1)</math> and <math>B(6)=A(6)</math> so |
− | + | :<math>A(1)\cdot A(6)+A(1)\cdot A(6)=\frac{5}{96}</math> | |
+ | :<math>A(1)\cdot A(6)=\frac{5}{192}</math> | ||
− | + | Also, we know that <math>A(2)=A(3)=A(4)=A(5)=\frac{1}{6}</math> and that the total probability must be <math>1</math>, so: | |
− | + | :<math>A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6}</math> | |
+ | :<math>A(1)+A(6)=\frac{1}{3}</math> | ||
− | + | Combining the equations: | |
− | + | :<math>A(6)\cdot(\frac{1}{3}-A(6))=\frac{5}{192}</math> | |
− | + | :<math>\frac{A(6)}{3}-A(6)^2=\frac{5}{192}</math> | |
− | <math> | + | :<math>A(6)^2-\frac{A(6)}{3}+\frac{5}{192}=0</math> |
− | <math> | + | :<math>192 A(6)^2 - 64 A(6) + 5 = 0</math> |
− | <math> | + | :<math>A(6)=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192}</math> |
− | + | :<math>A(6)=\frac{64\pm16}{384}</math> | |
− | <math> ( | + | :<math>A(6)=\frac{5}{24}, \frac{1}{8}</math> |
− | <math> | + | We know that <math>A(6)>\frac{1}{6}</math>, so it can't be <math>\frac{1}{8}</math>. Therefore, it has to be <math>\frac{5}{24}</math> and the answer is <math>5+24=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. | |
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== See also == | == See also == |
Revision as of 14:08, 25 September 2007
Problem
The number can be written as
where
and
are positive integers. Find
.
Solution
For now, 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
and
, 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, it has to be
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.
See also
2006 AIME I (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 |