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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+87309 
+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
 #1
avatar+87309 
+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+92805 
+5

Hi Chris,   

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

Melody  Apr 6, 2015
 #3
avatar+87309 
0

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

 

  

CPhill  Apr 6, 2015
 #4
avatar+92805 
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

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