Solution

Uniform area density in the unit disk is

1/ π

. The polar coordinates Jacobian gives the joint density

f_{R, Θ} (r, θ) = \frac{r}{π} = (2 r) (\frac{1}{2 π}), 0 \leq r \leq 1, 0 \leq θ < 2 π .

Thus the coordinates are independent and the uniform random point in a disk has

f_{R} (r) = 2 r 1_{[0, 1]} (r), f_{Θ} (θ) = \frac{1}{2 π} 1_{[0, 2 π)} (θ) .

For independent points

A, B

, the area of triangle

O A B

\frac{1}{2} R_{A} R_{B} ∣ sin (Θ_{A} - Θ_{B}) ∣.

Now

E R = \int_{0}^{1} 2 r^{2} d r = \frac{2}{3}, E ∣ sin (Θ_{A} - Θ_{B}) ∣ = \frac{2}{π} .

Independence therefore gives

E Area (O A B) = \frac{4}{9 π} .

Conditioned on

A, B

, the probability that

C

lies inside

O A B

is its area divided by the disk area

π

. Hence

P (C \in △ O A B) = \frac{4}{9 π ^{2}} .

Four points in general position fail to form a convex quadrilateral exactly when one lies inside the triangle formed by the other three. Each of

A, B, C

lies inside the triangle formed by

O

and the other two with probability

4/ (9 π^{2})

, and these three events are disjoint. The remaining possibility is that

O

lies inside

A BC

1/4

. Therefore

P (O, A, B, C form a convex quadrilateral) = 1 - \frac{1}{4} - \frac{4}{3 π ^{2}} = \frac{3}{4} - \frac{4}{3 π ^{2}} .

Solved by gpt-5.6-sol high.

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