Thread: please help me in solving this probability problem

in a convex octagon in which no three diagonals are concurrent,two diagonals are drawn at random,then find the probability that they intersect at an internal point

thanks

There are 5 diagonals from each vertex (a diagonal can be drawn from each vertex to each of the points except itself and its two "neighbors"- 8- 3= 5). That means there are a total of 8(5)= 40 diagonals. Given a first vertex, each of its diagonals cross all 5 of the diagonals from its two neighbor vertices. Each of its diagonals cross only 4 of the diagonals of the other 5 vertices since one of those diagonals is to the original vertex. That is, of the 40 total diagonals, the diagonals from any one vertex cross 2(5)+ 5(4)= 30 of them. Assuming all diagonals are equally likely, the probablility of crossing is 30/40= 3/4= 0.75.