\[\boxed{\mathbf{199.}}\]
\[Длина\ полуокружности:\]
\[L = \frac{\text{πR}}{2}.\]
\[Дуга\ AB = 180{^\circ}.\]
\[\textbf{а)}\ 90{^\circ} = \frac{1}{2} \cup AB;\]
\[P(X) = \frac{\frac{1}{2} \cdot \frac{\text{πR}}{2}}{\frac{\text{πR}}{2}} = \frac{1}{2} = 0,5.\]
\[\textbf{б)}\ \cup BC = 120{^\circ};\]
\[\cup AC = 180{^\circ} - 120{^\circ} = 60{^\circ};\]
\[P(X) = \frac{\frac{1}{3} \cdot \frac{\text{πR}}{2}}{\frac{\text{πR}}{2}} = \frac{1}{3}.\]
\[\textbf{в)}\ \cup CD = 60{^\circ} - 30{^\circ} = 30{^\circ};\]
\[30{^\circ} = \frac{1}{6} \cup AB;\]
\[P(X) = \frac{\frac{1}{6} \cdot \frac{\text{πR}}{2}}{\frac{\text{πR}}{2}} = \frac{1}{6}.\]