There are $5$ points in the plane no three of which are collinear. We draw all the segments whose vertices are these points. What is the minimum number of new points made by the intersection of the drawn segments?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 5$
1. **Understanding the Problem:**
   We are given 5 points in the plane, no three of which are collinear. We need to determine the minimum number of new points created by the intersection of the segments formed by these points.

2. **Counting the Segments:**
   The number of segments that can be drawn between 5 points is given by the combination formula:
   \[
   \binom{5}{2} = \frac{5!}{2!(5-2)!} = 10
   \]
   So, there are 10 segments.

3. **Analyzing Intersections:**
   Since no three points are collinear, any two segments that intersect must do so at a point that is not one of the original 5 points. We need to determine the minimum number of such intersection points.

4. **Drawing a Diagram:**
   Drawing a diagram helps visualize the problem. Consider a pentagon with its diagonals drawn. The diagonals intersect inside the pentagon. 

5. **Counting the Intersections:**
   - Each pair of diagonals intersects exactly once.
   - In a pentagon, there are 5 diagonals.
   - The intersections of these diagonals are the new points we are interested in.

6. **Calculating the Intersections:**
   - The number of intersections of the diagonals inside the pentagon can be calculated by considering the number of ways to choose 4 points out of 5 to form two intersecting diagonals:
   \[
   \binom{5}{4} = 5
   \]
   This means there are 5 new intersection points formed by the diagonals.

7. **Conclusion:**
   The minimum number of new points made by the intersection of the drawn segments is 5.

\[
\boxed{5}
\]