# Probability that a uniformly distributed point lies within a triangle in the unit disc

Question: Consider $$\|{z}\|\le1$$ in the complex plane (ie a disc radius 1 centred at the origin). The points X and Y are independently & uniformly distributed within the disc. A third point W is added (also uniform & independent)

a) what is the probability that W lies within the triangle OXY.

b) What is the probability that OXYW form a convex quadrilateral

My attempt: I converted into modulus argument form, and calculated the distributions for the modulus & arguments. My thoughts for part a is to calculate the area of the triangle, and divide by the area of the circle. My equation for area is $$A=\frac{1}{2}r_xr_y\sin\theta_y$$ (I assumed wlog $$\theta_x =0$$ due to rotational symmetry). I found the expectation of this (I’m not sure if that’s any use), I’m not really sure what else I can do.

For part b I thought that W must be in the same half circle as Y and X, and it cannot be within the triangle. So I would do $$\frac{\frac{\pi}{2}-A}{\pi}$$

• X and Y do not define a single half circle. If OX and OY and XY partition the circle into seven parts, then (a) corresponds to the triangle OXY while (b) corresponds to three of the remaining parts - the ones with four boundaries. Commented May 12 at 11:23
• @Henry thank you! I’ve drawn it out & got the parts now. Do you have any advice for finding the area of the segments? I know the area of the triangle from the first part, I need the other 3 ‘bad’ parts (or the 3 good parts). Commented May 13 at 8:52

a) the probability of W lying in the triangle is just $$\frac{E(Area_{triangle})}{Area_{circle}}$$
b) a quadrilateral is NOT convex if the fourth point lies within the triangle made by the remaining 3 points. There’s 4 different triangles that can be made (OXY OXW OYW and XYW). The probability of the 4th point lying in the first 3 triangles was calculated in part a. The probability of the origin lying in XYW can be calculated as $$\frac{1}{4}$$ (which is easy to see by drawing a diagram).