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?

Answer 1

The probability that no two ants will reach the same vertex is 5/256.

Plane, point B and attention. Ants that move to D can come from either E or C. Ants moving to E can come from either B or D of points A and F of the octahedron. Therefore, there are 2 degrees of freedom to decide which ants move to D, 2 degrees of freedom to decide which ants move to E, and 2 degrees of freedom to decide which ants move to A.

So there are 2× 2× 2=8 ways an ant can move to different points.

There are 8 ways for ants to move to different points.

Points B, there are a total of 8 + 4 + 8 = 16 ways Ant can move to her different points. Aligned the square so that point B is defined as point where the ant moved from point A.

An ant can actually move from point A to 4 different points, so there are a total of 4×20=80 ways an ant can move to different points.

Each ant acts independently and has 4 different points to choose from.

So each ant has a 1/4 chance of moving to the desired location. Since he has 6 ants, the probability of appearance is

= 1/4096

So, the desired answer is 80/4096

= 5/256

To learn more about probability, visit:

brainly.com/question/13024117

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