Difference between revisions of "2006 AMC 12A Problems/Problem 20"

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{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}}
 
{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #20]] and [[2006 AMC 10A Problems/Problem 25|2006 AMC 10A #25]]}}
 
== Problem ==
 
== Problem ==
A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the [[probability]] that after seven moves the bug will have visited every vertex exactly once?
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A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?
  
<math> \mathrm{(A) \ } \frac{1}{2187}\qquad \mathrm{(B) \ } \frac{1}{729}\qquad \mathrm{(C) \ } \frac{2}{243}\qquad \mathrm{(D) \ } \frac{1}{81} \qquad \mathrm{(E) \ }  \frac{5}{243}</math>
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<math> \textbf{(A) } \frac{1}{2187}\qquad \textbf{(B) } \frac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad \textbf{(E) }  \frac{5}{243}</math>
  
== Solutions ==
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==Solution 1==
===Solution 1===
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Call this cube <math>ABCDEFGH</math>, with <math>A</math> being the starting point.
Call this cube ABCDEFGH, with A being the starting point. Because in 7 moves the bug has to visit the other vertices, the bug cannot revisit any vertex. Therefore, starting at A, the bug has a 3/3 chance of finding a good path to the next vertex, and call it B. Then, the bug has a 2/3 chance of reaching a new vertex next. Call this C. A, B, and C are on the same plane always. Now, we split cases.  
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In the first case, the bug goes to the vertex E opposite A on the space diagonal with probability 1/3. Then, the bug has to visit D on the plane of ABC last, as there is no way in and out from D. Therefore, there is only 1 way out of 81 to get to D last. Therefore, there is a  
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Because in <math>7</math> moves the bug has to visit the other vertices, the bug cannot revisit any vertex.
<math>1/243 \cdot 6/9 = 6/2187</math> chance of finding a good path in this case.
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In the second case, the bug goes to vertex D on plane ABC with a chance of 1/3. The bug then has only 1 way to go to a point E on the opposite face, therefore having a 1/3 probability. Then, the bug has a choice of two vertices on the face opposite to ABCD. This results in a 2/3 probability of finding a good path to a point F. Then, there is only 1 way out of 9 to visit both other vertices on that face in 2 moves. Multiply the probabilities for this case to get <math>2/243 \cdot 6/9 = 12/2187</math>.
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Therefore, starting at <math>A</math>, the bug has a <math>\frac{3}{3}</math> chance of finding a good path to the next vertex, and call it <math>B</math>.
Add the probabilities of these two cases together to get <math>\frac{18}{2187} = \frac2{243}\Rightarrow \boxed{\mathrm (C)}</math>
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===Solution 2===
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Then, the bug has a <math>\frac{2}{3}</math> chance of reaching a new vertex next. Call this <math>C</math>. <math>A, B,</math> and <math>C</math> are always on the same plane.
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 +
Now, we split cases.
 +
 
 +
Case <math>1</math>: The bug goes to the vertex <math>E</math> opposite <math>A</math> on the space diagonal with probability <math>\frac{1}{3}</math>.
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Then, the bug has to visit <math>D</math> on the plane of <math>ABC</math> last, as there is no way in and out from <math>D</math>.
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 +
Therefore, there is only <math>1</math> way out of <math>81</math> to get to <math>D</math> last.
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 +
Therefore, there is a <math>\frac{1}{243} \cdot \frac{6}{9} = \frac{6}{2187}</math> chance of finding a good path in this case.
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 +
Case <math>2</math>: The bug goes to vertex <math>D</math> on plane <math>ABC</math> with a chance of <math>\frac{1}{3}</math>.
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The bug then has only <math>1</math> way to go to a point <math>E</math> on the opposite face, therefore having a <math>\frac{1}{3}</math> probability.
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Then, the bug has a choice of two vertices on the face opposite to <math>ABCD</math>.
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This results in a <math>\frac{2}{3}</math> probability of finding a good path to a point <math>F</math>.
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Then, there is only <math>1</math> way out of <math>9</math> to visit both other vertices on that face in <math>2</math> moves.
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Multiply the probabilities for this case to get <math>\frac{2}{243} \cdot \frac{6}{9} = \frac{12}{2187}</math>.
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 +
Add the probabilities of these two cases together to get <math>\frac{18}{2187} = \boxed{\textbf{(C) }\frac{2}{243}}.</math>
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==Solution 2==
  
  
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</asy>
 
</asy>
  
Let us count the good paths. The bug starts at an arbitrary [[vertex]], moves to a neighboring vertex (3 ways), and then to a new neighbor (2 more ways).  So, [[without loss of generality]], let the [[cube (geometry) | cube]] have vertices <math>ABCDEFGH</math> such that <math>ABCD</math> and <math>EFGH</math> are two opposite [[face]]s with <math>A</math> above <math>E</math> and <math>B</math> above <math>F</math>.  The bug starts at <math>A</math> and moves first to <math>B</math>, then to <math>C</math>.
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Let us count the good paths. The bug starts at an arbitrary [[vertex]], moves to a neighboring vertex (<math>3</math> ways), and then to a new neighbor (<math>2</math> more ways).  So, without loss of generality, let the cube have vertices <math>ABCDEFGH</math> such that <math>ABCD</math> and <math>EFGH</math> are two opposite faces with <math>A</math> above <math>E</math> and <math>B</math> above <math>F</math>.  The bug starts at <math>A</math> and moves first to <math>B</math>, then to <math>C</math>.
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 +
From this point, there are two cases.
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 +
Case <math>1</math>: the bug moves to <math>D</math>.  From <math>D</math>, there is only one good move available, to <math>H</math>.  From <math>H</math>, there are two ways to finish the trip, either by going <math>H \to G \to F \to E</math> or <math>H \to E \to F \to G</math>.  So there are <math>2</math> good paths in this case.
  
From this point, there are two [[casework|cases]].  
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Case <math>2</math>: the bug moves to <math>G</math>.
  
Case 1: the bug moves to <math>D</math>.  From <math>D</math>, there is only one good move available, to <math>H</math>.  From <math>H</math>, there are two ways to finish the trip, either by going <math>H \to G \to F \to E</math> or <math>H \to E \to F \to G</math>.  So there are 2 good paths in this case.
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Case <math>2a</math>: the bug moves <math>G \to H</math>.  In this case, there are <math>0</math> good paths because it will not be possible to visit both <math>D</math> and <math>E</math> without double-visiting some vertex.
  
Case 2: the bug moves to <math>G</math>.
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Case <math>2b</math>: the bug moves <math>G \to F</math>.  There is <math>1</math> good path in this case, <math>F \to E \to H \to D</math>.
Case 2a: the bug moves <math>G \to H</math>.  In this case, there are 0 good paths because it will not be possible to visit both <math>D</math> and <math>E</math> without double-visiting some vertex.
 
Case 2b: the bug moves <math>G \to F</math>.  There is a unique good path in this case, <math>F \to E \to H \to D</math>.
 
  
Thus, all told we have 3 good paths after the first two move, for a total of <math>3\cdot 3 \cdot 2 = 18</math> good paths.  There were <math>3^7 = 2187</math> possible paths the bug could have taken, so the [[probability]] a random path is good is the [[ratio]] of good paths to total paths, <math>\frac{18}{2187} = \frac2{243}\Rightarrow \boxed{\mathrm (C)}</math>.
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Thus, all told we have <math>3</math> good paths after the first two moves, for a total of <math>3\cdot 3 \cdot 2 = 18</math> good paths.  There were <math>3^7 = 2187</math> possible paths the bug could have taken, so the probability of a random path is good is <math>\frac{18}{2187} = \boxed{\textbf{(C) }\frac2{243}}</math>.
  
===Solution 2 (using the answer choices)===
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==Solution 3 (answer choices)==
As in Solution 1, the bug can move from its arbitrary starting vertex to a neighboring vertex in 3 ways. After this, the bug can move to a new neighbor in 2 ways (it cannot return to the first vertex). The total number of paths (as stated above) is <math>3^7</math> or <math>2187</math>. Therefore, the probability of the bug following a good path is equal to <math>\frac{6x}{2187}</math> for some positive integer <math>x</math>. The only answer choice which can be expressed in this form is <math>\frac2{243}\Rightarrow\boxed{\mathrm (C)}</math>.
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As in Solution 1, the bug can move from its arbitrary starting vertex to a neighboring vertex in <math>3</math> ways. After this, the bug can move to a new neighbor in <math>2</math> ways (it cannot return to the first vertex). The total number of paths (as stated above) is <math>3^7</math> or <math>2187</math>. Therefore, the probability of the bug following a good path is equal to <math>\frac{6x}{2187}</math> for some positive integer <math>x</math>. The only answer choice which can be expressed in this form is <math>\boxed{\textbf{(C) }\frac2{243}}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 10:29, 19 December 2021

The following problem is from both the 2006 AMC 12A #20 and 2006 AMC 10A #25, so both problems redirect to this page.

Problem

A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?

$\textbf{(A) } \frac{1}{2187}\qquad \textbf{(B) } \frac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad \textbf{(E) }  \frac{5}{243}$

Solution 1

Call this cube $ABCDEFGH$, with $A$ being the starting point.

Because in $7$ moves the bug has to visit the other vertices, the bug cannot revisit any vertex.

Therefore, starting at $A$, the bug has a $\frac{3}{3}$ chance of finding a good path to the next vertex, and call it $B$.

Then, the bug has a $\frac{2}{3}$ chance of reaching a new vertex next. Call this $C$. $A, B,$ and $C$ are always on the same plane.

Now, we split cases.

Case $1$: The bug goes to the vertex $E$ opposite $A$ on the space diagonal with probability $\frac{1}{3}$.

Then, the bug has to visit $D$ on the plane of $ABC$ last, as there is no way in and out from $D$.

Therefore, there is only $1$ way out of $81$ to get to $D$ last.

Therefore, there is a $\frac{1}{243} \cdot \frac{6}{9} = \frac{6}{2187}$ chance of finding a good path in this case.

Case $2$: The bug goes to vertex $D$ on plane $ABC$ with a chance of $\frac{1}{3}$.

The bug then has only $1$ way to go to a point $E$ on the opposite face, therefore having a $\frac{1}{3}$ probability.

Then, the bug has a choice of two vertices on the face opposite to $ABCD$.

This results in a $\frac{2}{3}$ probability of finding a good path to a point $F$.

Then, there is only $1$ way out of $9$ to visit both other vertices on that face in $2$ moves.

Multiply the probabilities for this case to get $\frac{2}{243} \cdot \frac{6}{9} = \frac{12}{2187}$.

Add the probabilities of these two cases together to get $\frac{18}{2187} = \boxed{\textbf{(C) }\frac{2}{243}}.$

Solution 2

[asy] import three; unitsize(1cm); size(50); currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); [/asy]

Let us count the good paths. The bug starts at an arbitrary vertex, moves to a neighboring vertex ($3$ ways), and then to a new neighbor ($2$ more ways). So, without loss of generality, let the cube have vertices $ABCDEFGH$ such that $ABCD$ and $EFGH$ are two opposite faces with $A$ above $E$ and $B$ above $F$. The bug starts at $A$ and moves first to $B$, then to $C$.

From this point, there are two cases.

Case $1$: the bug moves to $D$. From $D$, there is only one good move available, to $H$. From $H$, there are two ways to finish the trip, either by going $H \to G \to F \to E$ or $H \to E \to F \to G$. So there are $2$ good paths in this case.

Case $2$: the bug moves to $G$.

Case $2a$: the bug moves $G \to H$. In this case, there are $0$ good paths because it will not be possible to visit both $D$ and $E$ without double-visiting some vertex.

Case $2b$: the bug moves $G \to F$. There is $1$ good path in this case, $F \to E \to H \to D$.

Thus, all told we have $3$ good paths after the first two moves, for a total of $3\cdot 3 \cdot 2 = 18$ good paths. There were $3^7 = 2187$ possible paths the bug could have taken, so the probability of a random path is good is $\frac{18}{2187} = \boxed{\textbf{(C) }\frac2{243}}$.

Solution 3 (answer choices)

As in Solution 1, the bug can move from its arbitrary starting vertex to a neighboring vertex in $3$ ways. After this, the bug can move to a new neighbor in $2$ ways (it cannot return to the first vertex). The total number of paths (as stated above) is $3^7$ or $2187$. Therefore, the probability of the bug following a good path is equal to $\frac{6x}{2187}$ for some positive integer $x$. The only answer choice which can be expressed in this form is $\boxed{\textbf{(C) }\frac2{243}}$.

See also

2006 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2006 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 24
Followed by
Last Question
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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