If a dot is put in the incircle of an equilateral triangle without looking into it,
(a) Find the probability of the dot being inside the incircle
​(b) Find the probability of the dot being outside the incircle.

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{if} \mathrm{r} \mathrm{is} \mathrm{radius} \mathrm{of} \mathrm{incircle} \mathrm{and} \mathrm{s} \mathrm{is} \mathrm{semiperimeter} \mathrm{and} △ \mathrm{is} \mathrm{area} \\ \mathrm{of} \mathrm{triangle} , \mathrm{then} \\ \mathrm{r}=\frac{△}{\mathrm{s}} \\ =\frac{{\displaystyle \frac{\sqrt{3}}{4}}{\mathrm{a}}^2}{{\displaystyle \frac{3\mathrm{a}}{2}}} \\ \mathrm{r}=\frac{\mathrm{a}}{2\sqrt{3}} \\ \mathrm{Hence} \mathrm{area} \mathrm{of} \mathrm{incircle}={\mathrm{\pi r}}^2=\mathrm{\pi }\left(\frac{\mathrm{a}}{2\sqrt{3}}\right)=\frac{{\mathrm{\pi a}}^2}{12} \\ \mathrm{Hence} \\ \left(\mathrm{i}\right) \mathrm{Probability} \mathrm{of} \mathrm{the} \mathrm{dot} \mathrm{being} \mathrm{inside} \mathrm{the} \mathrm{incircle}=\frac{\mathrm{Area} \mathrm{of} \mathrm{incircle}}{△} \\ =\frac{\frac{{\mathrm{\pi a}}^2}{12}}{{\displaystyle \frac{\sqrt{3}}{4}}{\mathrm{a}}^2} \\ =\frac{\mathrm{\pi }}{3\sqrt{3}} \\ \left(\mathrm{ii}\right) \mathrm{Probability} \mathrm{of} \mathrm{the} \mathrm{dot} \mathrm{being} \mathrm{outside} \mathrm{the} \mathrm{incircle}= \\ 1- \mathrm{Probability} \mathrm{of} \mathrm{the} \mathrm{dot} \mathrm{being} \mathrm{inside} \mathrm{the} \mathrm{incircle} \\ =1-\frac{\mathrm{\pi }}{3\sqrt{3}} \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards