A) \[\frac{1}{4}\]
B) \[\frac{5}{12}\]
C) \[\frac{1}{16}\]
D) \[\frac{5}{16}\]
Correct Answer: A
Solution :
\[AB=\frac{5}{6}r\Rightarrow CD=\frac{2}{3}r\] \[\Rightarrow \,\,\,\,\,OB=\sqrt{{{r}^{2}}-\frac{25}{36}{{r}^{2}}}\] \[\Rightarrow \,\,\,\,\,OB=\sqrt{\frac{11}{6}}r\] |
Similarly, \[OD=\sqrt{{{r}^{2}}-\frac{4}{9}{{r}^{2}}}=\frac{\sqrt{5}}{3}r\] |
Required probability\[=\frac{\pi \left( \frac{5}{9}{{r}^{2}}-\frac{11}{36}{{x}^{2}} \right)}{\pi {{r}^{2}}}=\frac{20-11}{36}=\frac{1}{4}\] |
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