If \(r\) is the radius of the circle, then the side length of the triangle is \(2r\sqrt3\).
The area of the circle is \(\pi r^2\), while the area of the triangle is \(3r^2\sqrt3\).
So, the probability of selecting a random point in the white space is
\(1-\dfrac{\pi r^2}{3r^2\sqrt3}=1-\dfrac{\pi}{3\sqrt3}\approx0.395\approx40\%\)
