Question

A bee is moving in three-dimensional space. A fair six-sided die with faces labeled $A^+, A^-, B^+, B^-, C^+,$ and $C^-$ is rolled. Suppose the bee occupies the point $(a, b, c).$ If the die shows $A^+,$ then the bee moves to the point $(a + 1, b, c),$ and if the die shows $A^-,$ then the bee moves to the point $(a - 1, b, c).$ Analogous moves are made with the other four outcomes.

Suppose the bee starts at the point $(0, 0, 0)$ and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?

$\dfrac{1}{54}$

$\dfrac{7}{54}$

$\dfrac{1}{6}$

$\dfrac{5}{18}$

$\dfrac{2}{5}$

Solution:

Every roll moves the bee one unit along $\pm x, \pm y,$ or $\pm z,$ so there are $6^4 = 1296$ equally likely move sequences. A sequence works exactly when its four unit steps are four distinct edges of one unit cube, meaning the bee stays on a single cube and never repeats an edge. Enumerating these gives $168$ favorable sequences, so the probability is $\tfrac{168}{1296} = \tfrac{7}{54}.$ Therefore, the answer is B.

2024 AMC 10A Problem 24

Solution:

Problem 24 in Other Years