Intersection of Diagonals

The number of ways we can take a diagonal pair of a $100-gon$ $=(100C_{2}-100)C_{2}$
[the number of diagonals = $100C_{2}$ ]

Among the diagonals only those pair will intersect between the polygon which are created using $4$ different points.
[we can also create $2$ pair by taking $3$ points but they will not intersect in between]

The way to take a pair of lines using $4$ different points $=(100C_{2}*98C_{2})/(2!)$
[here we divided by $2!$ because we are taking $2$ groups with same size and we don't care about the serial of taking them, if they were of different size we would not have worried about it that is we would not have divided by $2!$ ]

But not all of them will intersect. We can create $3$ different pairs with these points and only 1 of them intersects.
So the number of such(correct) pairs $=(100C_{2}*98C_{2})/(2*3)$

So the probability = $\frac{(100C_{2}*98C_{2})/(2*3)}{(100C_{2}-100)C_{2}}$ = $\frac{3921225}{11758825}$ = $\frac{1617}{4849}$