= Solution
Uniform area density in the unit disk is $1/\pi$. The <polar coordinates> Jacobian gives the joint density
$$
f_{R,\Theta}(r,\theta)
=\frac r\pi
=\left(2r\right)\left(\frac1{2\pi}\right),
\qquad
0\leq r\leq1,\quad0\leq\theta<2\pi.
$$
Thus the coordinates are independent and the <uniform random point in a disk> has
$$
\boxed{
f_R(r)=2r\mathbf1_{[0,1]}(r),
\qquad
f_\Theta(\theta)=\frac1{2\pi}\mathbf1_{[0,2\pi)}(\theta)}.
$$
For independent points $A,B$, the area of triangle $OAB$ is
$$
\frac12R_AR_B|\sin(\Theta_A-\Theta_B)|.
$$
Now
$$
\mathbb ER=\int_0^1 2r^2\,dr=\frac23,
\qquad
\mathbb E|\sin(\Theta_A-\Theta_B)|=\frac2\pi.
$$
Independence therefore gives
$$
\boxed{\mathbb E\,\operatorname{Area}(OAB)=\frac4{9\pi}}.
$$
Conditioned on $A,B$, the probability that $C$ lies inside $OAB$ is its area divided by the disk area $\pi$. Hence
$$
\boxed{\mathbb P(C\in\triangle OAB)=\frac4{9\pi^2}}.
$$
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 $4/(9\pi^2)$, and these three events are disjoint. The remaining possibility is that $O$ lies inside $ABC$. By <origin in a triangle of three radial random points>, this has probability $1/4$. Therefore
$$
\boxed{
\mathbb P(O,A,B,C\text{ form a convex quadrilateral})
=1-\frac14-\frac{4}{3\pi^2}
=\frac34-\frac4{3\pi^2}}.
$$
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