Two points are uniformly and independently distributed (located) inside a square. A circle is drawn such that the segment joining the two points is a diameter. Find the probability that the center of the square lies inside that circle.
Call the first point A, the second point B, and the center O. Join the line AO, and extend a line DOE through O, perpendicular to AO and continuing in both directions. If B is on the other side of DOE from A, then the circle joining A and B will contain the center. If B is on the same side of the line as A, then it will not. The line bisects the square's area, so the probability is 1/2.
Although the answer has already been given, it's possible to check this via simulation. In R:
n = 10^8 x1 = runif(n, 0, 1) x2 = runif(n, 0, 1) y1 = runif(n, 0, 1) y2 = runif(n, 0, 1) x.center = (x1+x2)/2 y.center = (y1+y2)/2 radius = sqrt((x.center-x1)^2+(y.center-y1)^2) distance.to.center = sqrt((x.center-1/2)^2+(y.center-1/2)^2) sum(radius > distance.to.center)
This simulates drawing $10^8$ pairs of points
(x1, y1), (x2, y2), and finds the center and radius of the corresponding circles, and checks to see whether those circles contain the center of the square (1/2, 1/2). When I ran this I got 49995062, meaning that the probability can be estimated as 0.49995062. I'd say that's close to 1/2.