A disc of unit radius is tossed at random onto a large rectangular floor, which is tiled with 4x6 tiles. Find the probability that the disc is contained entirely in a rectangular tile (and does not intersect the border between two tiles).
We want the disc to have a distance of at least 1 from the sides, so the probability is (3*5)/(4*6) = 5/8.
Actually I belive it should be 1/3, because if the disc radius is 1, then you know that the center has to be contained within a 2x4 rectangle(since you subtract 1 of both dimensions of the 4x6 rectangle) That gives you a possible area for the center of the circle, retaining an area of 8, out of the entire area of 24, giving us 8/24 = 1/3
