Two of the vertices of a regular octahedron are to be chosen at random. What is the probability that they will be the endpoints of an edge of the octahedron? Express your answer as a common fraction.

Guest Apr 5, 2015

#1**+10 **

Here's a picture of a regular octahedron :

There are 6 vertices and the number of ways of pairing any two of them is C(6,2) = 15

But notice that only 12 pairs of them form edges. {The top and bottom vertices aren't connected to each other and neither are the two opposite vertices on the "sides." }

So, the probability that any two form an edge is 12/15 = 4/5

CPhill
Apr 6, 2015

#1**+10 **

Best Answer

Here's a picture of a regular octahedron :

There are 6 vertices and the number of ways of pairing any two of them is C(6,2) = 15

But notice that only 12 pairs of them form edges. {The top and bottom vertices aren't connected to each other and neither are the two opposite vertices on the "sides." }

So, the probability that any two form an edge is 12/15 = 4/5

CPhill
Apr 6, 2015