+0  
 
0
634
4
avatar

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

Best Answer 

 #1
avatar+85789 
+10

Here's a picture of a  regular octahedron :

Octahedron.jpg

 

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
Sort: 

4+0 Answers

 #1
avatar+85789 
+10
Best Answer

Here's a picture of a  regular octahedron :

Octahedron.jpg

 

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
 #2
avatar+92221 
+5

Hi Chris,   

That makes sense :)  I like your pic too :)

Melody  Apr 6, 2015
 #3
avatar+85789 
0

Thanks, Melody......I "stole" that pic from Wikipedia.....LOL!!!

 

  

CPhill  Apr 6, 2015
 #4
avatar+92221 
0

I knew that you got it from the web.  It is a good pic.  Pictures make answers so much easier to understand.

They also make answers look 'prettier' and 'pretty' answers are less intimidating to work through. :)

Melody  Apr 6, 2015

22 Online Users

avatar
avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details