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.

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Answer:

[tex]P_{bottom}=\frac{1}{4}=0.25[/tex]

Step-by-step explanation:

Starting from the top, the ant can only take four different directions, all of them going down, every direction has a probability of 1/4. For the second step, regardless of what direction the ant walked, it has 4 directions: going back (or up), to the sides (left or right) and down. If the probability of the first step is 1/4 for each direction and once the ant has moved one step, there are 4 directions with the same probability (1/4  again), the probability of taking a specific path is the multiplication of the probability of these two steps:

[tex]P_{2steps}=\frac{1}{4}*\frac{1}{4}=\frac{1}{16}[/tex]

There are only 4 roads that can take the ant to the bottom in 2 steps, each road with a probability of 1/16, adding the probability of these 4 roads:

[tex]P_{bottom}=\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}=\frac{4}{16}=\frac{1}{4}[/tex]

The probability of the ant ending up at the bottom is [tex]\frac{1}{4}[/tex] or 0.25.

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Answer:

[tex]\frac{1}{4}[/tex]

Explanation:

Ant has 4 ways to go at first. No matter where it goes each way has 4 more ways. So to get to the bottom from 4 ways, there is only one way so, 1/4 is the answer