△=4a^2-4(-b^2+π)=4(a^2+b^2-π)>=0,∴a^2+b^2>=π,设区域D={(a,b)|-π<=a<=π,-π<=b<=π,a^2+b^2>=π},则D的面积S1=4π^2-π^2=3π^2,正方形：“-π<=a<=π,-π<=b<=π”的面积S=4π^2。∴所求概率=S1/S=3/4.
(4*pi^2-p^3)/4*pi^2
f（x）=x^2+2ax-b^2+π有零点的概率为1/π
