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Difference between revisions of "2024 AMC 10A Problems/Problem 24"
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== Solution 1 ==
 
== Solution 1 ==
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[[User:Juwushu|Juwushu]] will finish this. WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a <math>\frac{6}{6}</math> probability of being along one of the 8 unit cubes around the origin, and the second move has a <math>\frac{4}{6}</math> chance.
+
WLOG, assume that the first two moves are equal for all possible combinations, since the direction does not matter. The first move has a <math>\frac{6}{6}</math> probability of being along one of the 8 unit cubes around the origin, and the second move has a <math>\frac{4}{6}</math> chance.
 
Now, there are two cases. We are currently on one of the points of the <math>2</math> by <math>2</math> squares that are aligned with the axes.
 
Now, there are two cases. We are currently on one of the points of the <math>2</math> by <math>2</math> 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 <math>\frac{2}{6}</math> chance of this happening and a <math>\frac{2}{6}</math> chance of the fourth move remaining on one of the cubes.
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The second case is if the bee moves along the same plane of the <math>2</math> by <math>2</math> squares previously, ending up on a point 1 away from the origin. There is a <math>\frac{1}{6}</math> chance of this happening and a <math>\frac{3}{6}</math> chance of remaining on one of the cubes.
 +
Now, multiply and sum for the answer.
 +
<cmath>\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})</cmath>
 +
Evaluating this gives you the answer of <math>\box{\textbf{(B) }\frac{7}{54}}</math>.
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Solution by [[User:Juwushu|juwushu]].

Revision as of 16:36, 8 November 2024

Problem

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?

$\textbf{(A) }\frac{1}{54}\qquad\textbf{(B) }\frac{7}{54}\qquad\textbf{(C) }\frac{1}{6}\qquad\textbf{(D) }\frac{5}{18}\qquad\textbf{(E) }\frac{2}{5}$

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 $\frac{6}{6}$ probability of being along one of the 8 unit cubes around the origin, and the second move has a $\frac{4}{6}$ chance. Now, there are two cases. We are currently on one of the points of the $2$ by $2$ 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 $\frac{2}{6}$ chance of this happening and a $\frac{2}{6}$ 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 $2$ by $2$ squares previously, ending up on a point 1 away from the origin. There is a $\frac{1}{6}$ chance of this happening and a $\frac{3}{6}$ chance of remaining on one of the cubes. Now, multiply and sum for the answer. \[\frac{2/3}\cdot(\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{6}\cdot\frac{1}{2})=\frac{2}{3}\cdot(\frac{1}{9}+\frac{1}{12})\] Evaluating this gives you the answer of $\box{\textbf{(B) }\frac{7}{54}}$ (Error compiling LaTeX. Unknown error_msg). Solution by juwushu.

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