Consider 2019 points chosen randomly and uniformly from the unit disc, defined as $\{(x, y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1\}$. We are interested in comparing the probabilities of the convex hull $C$ being a polygon with exactly three vertices (a triangle) versus exactly four vertices (a quadrilateral).

The key idea is to use geometric probability and properties of random points within a circle. The convex hull of points is most likely to form a polygon with $k$ vertices when the distribution of points near the boundary has more variability. The more points there are, the higher the probability that they are distributed along the boundary in such a way as to increase the number of vertices.

For a set of random points in a disc, the expected number of vertices of the convex hull is approximately $\log n$, where $n$ is the number of points. In our case, with 2019 points, this suggests a larger number of vertices on average than just three or four.

### Triangle Probability
The probability that exactly three points are on the boundary of the convex hull (forming a triangle) is low because:
1. The scenario requires every other point to be inside the triangle formed by the selected three boundary points.
2. As the number of points increases, it becomes less likely for the convex hull to be a simple triangle unless points are very specifically positioned.

### Quadrilateral Probability
The probability that exactly four points form the vertices of the convex hull (forming a quadrilateral) is somewhat higher because:
1. This arrangement allows for more combinations and slight adjustment of one of the vertices without completely altering the structure of the boundary.
2. With 2019 points, the chance of having exactly four distinct points on the boundary without any internal point aligning exactly with them typically increases.

Given the symmetry and placement randomness, combined with the number of points involved, the probability that the convex hull of the points forms a quadrilateral is higher than forming just a triangle.

Hence, the answer can be summarized as:
$$\boxed{\text{The probability that } C \text{ is a quadrilateral is larger than the probability that } C \text{ is a triangle.}}$$

This conclusion is based on the greater flexibility and higher likelihood of multiple boundary points forming a polygon with more than three vertices.