, 2012

Since $PQ$ passes through centre $O$, then it is a diameter of the circle. [[IMAGE0]] Since $\angle QOT=90\degree$, then $\angle POT=180\degree-90\degree=90\degree$. Thus, the area of sector $POT$ is $\frac{90\degree}{360\degree}=\frac{1}{4}$ or 25% of the area of the circle. Since the areas labelled $R$ and $S$ are equal, then each is $25\%\div 2=12.5\%$ of the area of the circle. Therefore, a spin will stop on the shaded region $12.5\%$ of the time.