Now I got it. This question was haunting me for last 24 hours. Atlast I nailed it.
First of all. If all three points are in the semicirle then the triangle will be obtuse.
Ways of selecting first point = 12
Ways of selecting 2nd point = 10 (You can select any point except the first and its opposite, for example if 12 selected then you can select any point other than 12 and 6)
Now comes the tricky part.
Average ways of selecting 3rd point is dependent on the second selection = 2 * (8+7+6+5+4) = 60/10 = 6
But there will be 5 more repeats of each combination of 3 points.
Total ways of selecting 3 points = 12C3 = 220
Prob = 12 * 10 * 6/220*6 = 6/11
We will take an example.
Let first selection is 12
1. Second selection is 1:
Third could be anything except these: opposite of 12 and 1 (i.e 6 and 7) and anything between them 6 and 7 (Nothing here) = 2,3,4,5,11,10,9,8 = 8 ways
2. Second selection is 2:
Third could be anything except these: opposite of 12 and 2 (i.e 6 and 8) and anything between 6 and 8 (i.e 7) = 1,3,4,5,11,10,9 = 7 ways
3. Second selection is 3:
Third could be anything except these: opposite of 12 and 3 (i.e 6 and 9) and anything between 6 and 9 (i.e 6,7,8,9) = 1,2,3,5,11,10 = 6 ways
4. Second selection is 4:
Third could be anything except these: opposite of 12 and 4 (i.e 6 and 10) and anything between 6 and 10 (i.e 6,7,8,9,10) = 1,2,3,5,11 = 5 ways
5. Second selection is 5:
Third could be anything except these: opposite of 12 and 11 (i.e 6 and 11) and anything between 6 and 11 (i.e 6,7,8,9,10,11) = 1,2,3,4 = 4 ways
Same way on the opposite side i.e when second selection is 11,10,9,8,7
Total average of selecting second and third = 2*(8+7+6+5+4) = 60
Total obtuse triangles = 12*60 = 720
But there are duplicates:
(12,1,2) there are 3! of duplicates of this.
So final prob = 12*60/(220*6) = 6/11
Kevin, your questions, especially this one, are too difficult to solve in 2 minutes.
_________________
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008