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 $3/2.$
Let $O$ be the center of the circle. Suppose the first point we choose is $A.$ Let $B$ be the second point we choose.
[asy]
unitsize(2 cm);
pair A, B, O;
O = (0,0);
A = dir(110);
B = dir(60);
draw(Circle(O,1));
draw((-1.5,0)--(1.5,0));
draw((0,-1.5)--(0,1.5));
draw(O--A);
draw(O--B);
label("$O$", O, SW);
label("$A$", A, NW);
label("$B$", B, N);
[/asy]
Then the distance $AB$ is at most $\frac{3}{2}$ if and only if $B$ lies inside the shaded annulus centered at $O.$
[asy]
unitsize(2 cm);
pair A, B, O;
O = (0,0);
A = dir(110);
B = (1.5,0);
draw(Circle(O,1));
draw((-1.5,0)--(1.5,0));
draw((0,-1.5)--(0,1.5));
draw(arc(O,3/2,0,180),red);
label("$O$", O, SW);
filldraw(arc(O,1,0,180)--(0,0)--cycle,gray(0.7));
label("$A$", A, N);
label("$B$", B, N);
[/asy]
The probability that the second point we choose lies inside this annulus is the ratio of the areas of the annulus and the whole circle, which is
\[\frac{\pi \left( \left( \frac{3}{2} \right)^2 - 1^2 \right)}{\pi (1)^2} = \boxed{\frac{5}{4}}.\]
To solve this problem, we can consider the set of all possible pairs of points on the circle. Let's label these points $A$ and $B.$
First, let's determine the total number of possible pairs of points. Since there are infinitely many points on the circle, we cannot directly count all the possible pairs. However, we can determine the number of possible pairs within a certain angle range.
Let's consider an angle range of $60^\circ.$ Using this angle, we can divide the circle into six equal sectors. Any pair of points on the circle will fall within one of these sectors.
Since each sector covers $60^\circ$ out of $360^\circ,$ the total angle covered by all six sectors is $6 \cdot 60^\circ = 360^\circ.$ This means that all possible pairs of points on the circle can be found in these six sectors.
Now, let's determine the total length of the circumference covered by these six sectors. The circumference of a circle with radius $1$ is $2\pi.$ Since the six sectors cover the entire circle, the total length of the circumference covered by these sectors is $2\pi.$
To find the probability that the distance between the points $A$ and $B$ is at most $3/2,$ we must find the probability that the pair $(A,B)$ falls within a certain distance on the circle.
The distance between two points on a circle is given by the length of the arc that connects them. If this arc is less than or equal to $3/2,$ then the distance between the points $A$ and $B$ is also at most $3/2.$
To calculate this probability, we need to find the length of the arcs that correspond to a distance of $3/2$ on the circumference of the circle.
The circumference of the circle is $2\pi,$ so the length of the entire circumference corresponds to a distance of $2\pi.$ Therefore, the length of the arc corresponding to a distance of $3/2$ is given by $\frac{3/2}{2\pi} \times 2\pi = 3.$
Since the length of the arc corresponding to a distance of $3/2$ is $3,$ the probability that a randomly chosen pair of points $A$ and $B$ on the circle has a distance at most $3/2$ is $\frac{3}{2\pi}.$
Therefore, the probability that the distance between the two points is at most $3/2$ is $\boxed{\frac{3}{2\pi}}.$
Let's consider the situation where we have chosen one point $A$ on the circle. We want to find the probability that the second point $B$ we choose is within a distance of $3/2$ of $A.$
[asy]
unitsize(2 cm);
pair A, B, O, C;
O = (0,0);
A = dir(60);
B = dir(110);
C = dir(150);
draw(Circle(O,1));
draw((0,-1.5)--(0,1.5));
draw((1.5,0)--(-1.5,0));
draw(O--A--(1.5*dir(60)));
draw(O--B--(1.5*dir(110)));
label("$A$", A, dir(45));
label("$B$", B, N);
label("$O$", O, SW);
[/asy]
Let $O$ be the center of the circle. Triangle $OAB$ in the diagram above is an isosceles triangle (since $OA = OB = 1$). Let $\theta$ be the degree measure of $\angle AOB.$
The distance between $A$ and $B$ is less than or equal to $3/2$ if and only if $\angle AOB \le 90^\circ.$ Also, the region where $\angle AOB \le 90^\circ$ for a given point $A$ is a $90^\circ$ sector of the circle centered at $A,$ with radius $3/2.$
[asy]
unitsize(2 cm);
draw(Circle((0,0),1));
draw(dir(30)--dir(150)--dir(270)--cycle);
draw(dir(210)--dir(330)--(1,0));
label("$90^\circ$", (0.3,0.1));
[/asy]
Thus, the answer to the problem is the average value of the probability that $B$ lies in the $90^\circ$ sector centered at $A$.
Consider a small arc on the circle that subtends a central angle of $d \theta.$ The angle of the $90^\circ$ sector that is inside this small arc (when a point $A$ is chosen at random) is $\frac{90^\circ}{360^\circ} d \theta = \frac{d \theta}{4}.$ The measure of this small arc is $dr,$ where $r = 1$ is the radius of the circle. Hence, the area of this small sector is
\[\frac{1}{2} \left( \frac{d \theta}{4} \right) \left( r^2 \right) = \frac{1}{8} r^2 d \theta = \frac{1}{8} d \theta.\]Because the circle has radius 1, the total probability is
\[\frac{1}{8} \int_{-\pi}^{\pi} d \theta = \frac{1}{8} \left[ \theta \right]_{-\pi}^{\pi} = \frac{1}{8} (2 \pi - (-2 \pi)) = \boxed{\frac{1}{4}}.\]