Question

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?

$\dfrac{1}{2187}$

$\dfrac{1}{729}$

$\dfrac{2}{243}$

$\dfrac{1}{81}$

$\dfrac{5}{243}$

Answer: C

Concepts:basic probability graph theory casework

Difficulty rating: 2070

Solution:

From the start there are $3^7$ equally likely $7$ -move walks. For a walk visiting all $8$ vertices, there are $3$ choices for the first move and $2$ for the second, since it cannot return to the starting vertex.

Label cube vertices by three-bit strings. By symmetry, after fixing those first two moves we may take the first three vertices to be $000,001,011.$ A branch check gives exactly these three completions: $\begin{aligned} &010,110,111,101,100,\\ &010,110,100,101,111,\\ &111,101,100,110,010. \end{aligned}$ Thus there are $3 \cdot 2 \cdot 3 = 18$ such walks.

The probability is $\dfrac{18}{3^7} = \dfrac{18}{2187} = \dfrac{2}{243}.$

Thus, the correct answer is C.

2006 AMC 12A Problem 20

Answer: C

Difficulty rating: 2070

Solution:

Problem 20 in Other Years