We have $\binom{\binom{10}{2}}{2}=990$ options to choose two segments. Now, let's see how many of these pairs intersect at some point. We have $2$ different situations. Number of pairs intersecting on the circumference: We have $10$ options to choose the endpoint that the pair intersect. So, we have $9$ points left and we should choose two of them. Hence, we have $10\cdot \dbinom{9}{2}=360$ pairs of segments that intersect on the circumference. Number of pairs intersecting inside the circle: Each $4$ different points form exactly $1$ pair of segments that intersect inside the circle. Thus, we have $\dbinom{10}{4}=210$ such pairs. In total, we have $360+210=570$ pairs that intersect at some point. The total number of pairs is $990$. Hence, the probability is $\dfrac{570}{990}=\frac{19}{33}$.

If we choose four points on the circumference, there are exactly one form two draw 2 segments that cut internally and the other two forms to draw 2 segments doesn't cut. Also we count each pair of segments once. We conclude that the probability is 1/3