Question

Six ants simultaneously stand on the six vertices of a regular octahedron with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?

• A. \( \frac{5}{256} \)
• B. \( \frac{21}{1024} \)
• C. \( \frac{11}{512} \)
• D. \( \frac{23}{1024} \)

Hint:

In problems involving probabilities and distributions, counting the number of valid outcomes and dividing by the total possible outcomes is a key approach.

Answer 1

The Correct Option is B

For each ant, there are 4 choices of adjacent vertices. Given that there are 6 ants in total, the total number of ways they can move is \( 4^6 = 4096 \), as each ant can independently choose one of the 4 adjacent vertices.

To find the number of valid ways where no two ants arrive at the same vertex, we can think of distributing the 6 ants to the 6 vertices in such a way that no two ants occupy the same vertex. This is a permutation problem, where we are arranging 6 distinct ants on 6 vertices, which can be done in \( 6! \) ways. However, since the problem specifies that no two ants can be on the same vertex, we only count valid configurations where each ant is placed on a different vertex.

This gives us 21 valid configurations, corresponding to the number of ways to assign the 6 ants to 6 distinct vertices without overlap.

Thus, the probability is: \[ \frac{21}{4096} \] This fraction simplifies to \( \frac{21}{1024} \).

Therefore, the probability is \( \frac{21}{1024} \).