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Fact: The probability two sectors with central angles $\alpha$ and $\beta$ intersect is $\dfrac{\alpha+\beta}{2\pi}$ .

This can easily be proven with a diagram. The second sector can be right to the left or right of the first, or anywhere in between. This results in an angle of $\alpha+\beta$ .

Now, to the problem. Fix the sector with central angle $\dfrac{\pi}{10}$ , and call it A. Consider the sector with central angle $\dfrac{2\pi}{10}$ , and call it B. The third sector is C. The probability B intersects A is $\dfrac{3}{20}$ , by the fact above.

Now, let $f$ be the function that models the intersection of A and B for each position. Since area is linearly dependent on angle, $f$ first rises, stays constant, and then falls (draw it yourself; draw a circle, a sector, and see where the second sector can intersect the first). The graph looks like a hill.

Either way, the point is that the expected value for the area of the intersection is $\dfrac{\pi}{15}$ . Now, by our fact above, the probability that C intersects our current intersection is $\dfrac{11}{60}$ . Once again, this is valid because area is linearly dependent on angle.

Our probability is, multiplying our two probabilities: $\dfrac{3}{20}\cdot\dfrac{11}{60}=\dfrac{11}{400}$ The answer is $\boxed{411}$ .

Case 1: The $\frac{2\pi}{10}$ -sector is totally inside the $\frac{3\pi}{10}$ -sector, the probability of that case is $\frac{\frac{3\pi}{10}-\frac{2\pi}{10}}{2\pi}=\frac{1}{20}$

Now, in order to make 3 sectors have a point in common, we only need to make the $\frac{\pi}{10}$ -sector have a point in common with the $\frac{2\pi}{10}$ -sector, the probability is $\frac{\frac{\pi}{10}+\frac{2\pi}{10}}{2\pi}=\frac{3}{20}$

The probability overall of that case is: $\frac{3}{20}.\frac{1}{20}=\frac{3}{400}$ .

Case 2: The $\frac{2\pi}{10}$ -sector and $\frac{3\pi}{10}$ -sector have a $\phi$ -sector in common with $0< \phi < \frac{2\pi}{10}$ . The probability of that case is $2. \frac{\frac{2\pi}{10}}{2\pi}=\frac{1}{5}$ .

Now, in order to make 3 sectors have a point in common, we only need to make the $\frac{\pi}{10}$ -sector have a point in common with the $\phi$ -sector, the probability is $\frac{\frac{\pi}{10}+\phi}{2\pi}$ .

Because $\phi$ is randomly from 0 to $\frac{2\pi}{10}$ , the average number is $\frac{\pi}{10}$ , the average probability is $\frac{\frac{\pi}{10}+\frac{\pi}{10}}{2\pi}=\frac{1}{10}$ .

The probability overall of that case is: $\frac{1}{5}.\frac{1}{10}=\frac{1}{50}$ .

Overall, the probability that 3 sectors have a point in common is $\frac{1}{50}+\frac{3}{400}=\frac{11}{400}$ .

The answer is 11+400=411.