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?

$\dfrac{1}{2187}$

$\dfrac{1}{729}$

$\dfrac{2}{243}$

$\dfrac{1}{81}$

$\dfrac{5}{243}$

Solution:

From the start there are $3^7$ equally likely $7$ -move walks. Consider a walk visiting all $8$ vertices: there are $3$ choices for the first move and $2$ for the second.

Labeling the first three vertices $A, B, C,$ the bug must next move to one of two vertices, and in each case the remaining moves are forced. This gives $3 \cdot 2 \cdot 3 = 18$ 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

Solution:

Problem 20 in Other Years