An ant begins at the top of the pictured octahedron. If the ant takes two "steps", what is the probability it ends up at the bottom of the octahedron? Assume a "step" is a journey from one vertex to an adjacent vertex along an edge.
Verified Answer --- 1/4
Solution:
The ant can only move along adjacent edges, so he only has 4 ways to get to the bottom. He goes from the top to any vertex of the square, and then he moves to the very bottom vertex. Then, we have to find the total number of ways he can move in 2 steps. We already have the 4 that are correct, so don't forget them. The ant can move to any vertex on the square, and then move to the side both ways. He can do this on any edge, giving us 2*4 = 8 total ways to move on the edges of the square. Additionally, he can go down from the top, and then return to the top. this gives us 4 more ways. Adding these all together, we get
4 + 8 + 4 = 16. If we put our correct solutions over our total ways, we get 1/4.
clearly one step will take the ant to the equator of the octahedron.
From there, there are 4 ways to go, only one of which continues on to the bottom.
So, p = 1/4
The answer is 1/4 because the ant has 4 different ways to move the first step and 4 different ways to move the second step. This makes a grand total of 16 steps. Out of these only 4 will get the ant to its destination: the bottom of the octahedron. So, the probability is 4/16 = 1/4.
Well, let me ask you this: Have you ever seen an ant using Google Maps? They have small brains and shaky little ant feet—getting directions would be a six-legged calamity! But since ants operate purely on instinct, the probability should largely depend on what mischief FedEx delivery guy has been causing at your top floor dumpsters.
To find the probability that the ant ends up at the bottom of the octahedron after taking two steps, we need to first determine the number of possible paths that the ant can take.
Let's break down the problem into smaller steps:
Step 1: From the starting vertex at the top of the octahedron, the ant can take 4 different edges to reach the 4 adjacent vertices. This means there are 4 possible paths for the ant after the first step.
Step 2: From each of the 4 adjacent vertices, the ant can again take 4 different edges to reach 4 other vertices. This gives a total of 4 * 4 = 16 possible paths for the ant after the second step.
Now, we need to find the number of paths that lead to the bottom of the octahedron:
Step 3: Among the 16 possible paths after the second step, we can see that only 2 of them can lead directly to the bottom vertex. These are the paths that either start from the top and take an edge downwards or start from the bottom and take an edge upwards.
Therefore, the number of paths that lead to the bottom of the octahedron is 2.
Finally, to find the probability, we divide the number of favorable outcomes (2) by the total number of possible outcomes (16):
P(ends up at the bottom) = 2 / 16 = 1/8
So, the probability that the ant ends up at the bottom of the octahedron after taking two steps is 1/8.
2/3
An ant begins at the top of the pictured octahedron. If the ant takes two "steps", what is the probability it ends up at the bottom of the octahedron? Assume a "step" is a journey from one vertex to an adjacent vertex along an edge.
[asy]
import three;
triple A=(0,0,0),B=(2,0,0),C=(2,2,0),D=(0,2,0), EE = (1,1,2), F = (1,1,-2);
currentprojection=perspective(1/2,-1,1/2);
draw(A--B--C);
dot(EE);
draw(D--A,dotted);
draw(D--EE,dotted);
draw(C--D,dotted);
draw(A--EE--B);
draw(C--EE);
draw(A--F--B);
draw(C--F--D,dotted);
[/asy]