Find the probability of points lying in the space shown below 
In the above figure, an equilateral triangle with sides of length l is shown. The edges of the triangle extrapolates beyond the vertices to create 3 regions of P and Q. O is the origin. The regions P and Q are bounded by a circle of radius r with its centre at the origin.
If n points are uniformly randomly scattered on the area of the circle, what is the probability that:


*

*At least one point lies in one of the region $P$?

*At least one point lies in each of the region marked as $P$?

*No Point lies in one of the region $P$?.


What is the relation between area of $P$ and $Q$?
 A: Assuming the center of the circle is the centroid of the triangle. I calculate the area of one of the three regions $P$ as the sum of a circular segment and an equilateral triangle's area. The equilateral side length I compute to be $s=(r^2-3l^2/16)^{1/2}-l/2$ and area $s^2\sqrt{3}/4$. To compute the area of the circular segment I determine the angle $\theta$ at the center subtending the segment as $\sin(\theta/2)=s/(2r)$, and then the area of the segment itself is the area of the sector subtended by $\theta$, $r^2\theta/2$, less the area of the subtended isosceles triangle, $r^2\sin\theta/2$. So the area of $P$ is 3 times
$$
\lambda(P)/3=s^2\sqrt{3}/4+r^2/2(\theta-\sin\theta),
$$
with $s,\theta$ as above.
Now, assuming independent distributions. 1. The probability "At least one point lies in one of the region $P$" I interpret to mean all $n$ points don't lie in the complement of $P$, which has probability $1-(\pi r^2-\lambda(P))^n.$ 2. For the second probability, use the inclusion-exclusion principle on the complementary event that any one of the 3 regions is empty. 3. This is just the complement of 1, as I interpreted it.
