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