SOLUTION: 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 edge

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Question 987391: 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?
I asked this question on another site but I do not understand. Someone was speaking about symmetry but I don't see it. I do know that the denominator is 2187 because of 3 possibilties for every move. Can someone really explain this interms of how to find the successive outcomes?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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?
I asked this question on another site but I do not understand. Someone was speaking about symmetry but I don't see it. I do know that the denominator is 2187 because of 3 possibilties for every move. Can someone really explain this interms of how to find the successive outcomes?
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Find a cube that you can hold in your hand (Rubic or sugar or something) sol
you can see the symmetry.
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Note:: You start a one of the 8 vertices. There are 7 other vertices
P(visit all 7 of those vertices) = 1 - P(you don't visit all 7)
Since there are 3 choices when you are at each vertex the probability
you go to one of those three is 1/3.
Ans:: P(visit all 7) = 1 - (1/3)^7 = 1 - (1/3)^7 = 2186/2187
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Cheers,
Stan H.
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