2024 AMC 10A Problems/Problem 24
Problem
A bee is moving in three-dimensional space. A fair six-sided die with faces labeled 
and 
is rolled. Suppose the bee occupies the point 
If the die shows 
, then the bee moves to the point 
and if the die shows 
then the bee moves to the point 
Analogous moves are made with the other four outcomes. Suppose the bee starts at the point 
and the die is rolled four times. What is the probability that the bee traverses four distinct edges of some unit cube?
Solution 1
WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a 
probability of being along one of the 8 unit cubes around the origin, and the second move has a 
chance.
Now, there are two cases. We are currently on one of the points of the 
by squares that are aligned with the axes.
The first case is if the bee moves to the corner of a cube farthest away from the origin.
Here, there is a 
chance of this happening and a chance of the fourth move remaining on one of the cubes.
The second case is if the bee moves along the same plane of the by squares previously, ending up on a point 1 away from the origin. There is a 
chance of this happening and a 
chance of remaining on one of the cubes.
Now, multiply and sum for the answer.
![\[\frac{2}{3}\cdot(\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{2})=\frac{2}{3}\cdot(\frac{1}{9}+\frac{1}{12})\]](http://latex.artofproblemsolving.com/2/3/1/231b6abb32e723df5468e5bd994af575c55d6dae.png)
Evaluating this gives you the answer of 
.
Solution by juwushu.
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
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| All AMC 10 Problems and Solutions | ||
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