Probability Question - please help! (1 Viewer)

[rebemol]

Could someone explain how to do this question? Thanks heaps!

Nine points lie inside a circle. No three of the points are collinear. Five of the points lie in sector 1, three lie in sector 2, and the other point lies in sector 3.
(i) Show that 84 triangles can be made using these points as vertices.

One triangle is chosen at random from all the possible triangles.
(i) Find the probability that the vertices of the triangle chosen lie one in each sector
(ii) Find the probability that the vertices of the triangle chosen lie all in the same sector.

 

[shafqat]

3 points to make a triangle, chosen from 9 possible points, so total number of possibilities is 9 choose 3 = 84.
The number of triangles with a vertice in each sector is 5 * 3 * 1 = 15. so prob = 15/84.

For the last one, for all points to be in one sector, they ccan be in sector 1 ( 5 choose 3 possibilities) or sector 2 (3 choose 3) = 11. so prob = 11/84

 

[rebemol]

wow..it all makes sense now.
Thanks heaps!!!