This is quite a nice problem for demonstrating hit-and-miss Monte-Carlo. We generate a large sample of points in the unit cube count number of cases where discriminant is positive and divide by the sample size.
Originally Posted by dely84
So the required probability is ~=0.255 or 25.5%
>NN=10000000; ..sample size
>xx=random(3,NN); ..sample of size NN of coefficients
>pp=sum(xx(1,: )^2>4*xx(2,: )*xx(3,: ))/NN ..count the positive disc and divide by sample size
>SE=sqrt((NN*pp*(1-pp)))/NN ..approx standard error of the estimate