+0

Probability

-1
106
7

Consider the regular octahedron as shown. Each edge of the octahedron has a length of 1. An ant starts at the vertex A and crawls a total distance of 3 units along the edges of the octahedron. Any time the ant reaches a vertex of the octahedron, it randomly chooses an edge to next crawl on that is different from the edge it just left. One such path the ant may take is shown.  What is the probability that the ant will end up back at point A ? Express your answer as a common fraction.

Jul 20, 2022

#1
+2448
0

There are $$4 \times 3 \times 3 = 24$$ paths you can take.

Note that the only way to end up at point A is if you go along 1 of the triangles.

There are 4 triangles that contain point A, but there are 2 ways to go around each, making for $$4 \times 2 = 8$$ successful paths.

This means that the probability is $${8 \over 24} = \color{brown}\boxed{1 \over 3}$$

Jul 20, 2022
#2
+118133
0

0

It cannot land back at point A, not after 3 moves

Jul 21, 2022
#3
+2448
0

This is what a path would look like:

BuilderBoi  Jul 21, 2022
#4
+118133
0

Your right Builderboi, I was thinking of an octagon.

Melody  Jul 22, 2022
#5
+118133
+1

1 * 0.5 * 0.25 = 0.125

Check:

Number of paths = 4^3 = 64

Number of favourable pths = 4*2*1 = 8

8/64 = 1/8  = 0.125

Jul 22, 2022
#6
+2448
-1

It can't go right back to the vertex it just left...

BuilderBoi  Jul 22, 2022
#7
+118133
0

ok, we have interpreted the question differently.

Your interpretation is more likely what is meant.

Melody  Jul 23, 2022