**Part 1**
$d^2 + \left( \dfrac{r}{2} \right)^2 = r^2$

$d = \sqrt{r^2 - \dfrac{r^2}{4}}$

$d = \dfrac{\sqrt{3}r}{2}$

$P = \dfrac{d}{r} = \dfrac{\sqrt{3}r/2}{r}$

$P = \dfrac{\sqrt{3}}{2} \approx 0.866$ ← [ C ] *answer for part 1*

**Part 2**

$P = \dfrac{\pi d^2}{\pi r^2} = \dfrac{(\sqrt{3}r/2)^2}{r^2}$

$P = \dfrac{3}{4} = 0.75$ ← [ D ] *answer for part 2*

**Part 3**

If one end of the chord is at *A*, the other end must be on arc *BDC*.

$P = \dfrac{4}{6}$

$P = 0.667$ ← [ B ] *answer for part 3*