Two points on a circle of radius 1 are chosen at random. Find the probability that the distance between the two points is at most 1.5.
Consider a circle of radius 1 with center O, and let A and B be two points on the circle. Then the distance between A and B is given by AB = 2sin(x), where x is the angle between OA and OB.
[asy]
pair O, A, B;
O = (0,0);
A = dir(240);
B = dir(30);
draw(circle(O,1));
draw(O--A--B--O);
label("$O$", O, NE);
label("$A$", A, SW);
label("$B$", B, NE);
label("$x$", O--A, S);
label("$x$", A--B, SE);
[/asy]
Since the radius of the circle is 1, we have that $0\leq x\leq \pi$.
The distance between A and B is at most 1.5 if and only if $2\sin x\leq 1.5$, i.e. $\sin x\leq 0.75$. Using the fact that $\sin$ is a decreasing function on the interval $[0,\pi]$, we see that this happens if and only if $0\leq x\leq \sin^{-1}(0.75)\approx 0.848$. Therefore, the probability that the distance between the two points is at most 1.5 is given by the ratio of the area of the sector of the circle corresponding to $0\leq x\leq \sin^{-1}(0.75)$ to the area of the whole circle. Thus,
$$P(\text{distance} \leq 1.5) = \frac{\text{Area of sector}}{\text{Area of circle}} = \frac{\frac{1}{2}(0.848)(1)^2}{\pi(1)^2} \approx 0.269$$
Therefore, the probability that the distance between the two points is at most 1.5 is approximately $\boxed{0.269}$.