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2 distinct vertices of an octagon are chosen at random. What is the probability that there is exactly one other vertex between them?

 Nov 12, 2020
 #1
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Here's my  best  attempt :

 

Call  the vertices  (A,B,C,D,E,F,G,H)

 

The  number of possible  combos of choosing any two vertices  (without regard to order)  =  8C2  = 28

 

Note  that  the  vertices chosen - again disregarding order - where there  is only one  vertex bt\etween them  are :

 

ABC , BCD, CDE, DEF, EFG, FGH, GHA, HAB   =  

 

AC, BD, CE, DF, EG, FH , GA, HB

 

So....the probability =  8/28 =   2/7

 

cool cool cool

 Nov 12, 2020

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