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 $\dfrac{570}{990}=\frac{19}{33}$.