I didn't have any idea how to do this until I read asinus's answer!
But after I saw his I wanted to try it.
Here, A is the center of the circle with a radius of 1.
probability that X ≤ 1 unit away from A = \(\frac{\text{area of blue sector}}{\text{area of triangle}}\)
\(\frac{\text{area of blue sector}}{\pi*1^2}=\frac{60^{\circ}}{360^{\circ}} \rightarrow \text{area of blue sector}=\frac{\pi}{6}\)
\(\text{area of triangle}=(2)*(\frac12)*(\frac32)*(\frac{3\sqrt3}{2}) =\frac{9\sqrt3}{4}\)
\(\text{probability}=\frac{\frac{\pi}{6}}{\frac{9\sqrt3}{4}}=\frac{\pi}6\cdot\frac4{9\sqrt3}=\frac{4\pi}{54\sqrt3}=\frac{2\sqrt3\pi}{81}\)
My answer is a little different because I put the area of the whole triangle in the denominator, but asinus put the area of the rest of the triangle, not including the sector, in the denominator. I don't really know for sure which is correct.
