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Four points are chosen at random from the interior of a circle. What is the probability that the sum of each of their distances from the center is greater than half the circumference of the circle?

I don't even know how to begin this problem. Any hints would be greatly appreciated.

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    $\begingroup$ The interior of a circle/sphere is empty. You probably mean disk. $\endgroup$ – drhab Aug 14 '18 at 9:39
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    $\begingroup$ @drhab: There are different (but related) concepts of “interior”. Lkryat speaks about the interior of the circle as a Jordan curve (see Wikipedia), not about the interior of the circle as a set. $\endgroup$ – celtschk Aug 14 '18 at 9:59
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    $\begingroup$ @celtschk I don't think so. If so then there would be no problem at all, since in that case all chosen points have distance $r$ from the center and $4r>\pi r$ is immediate. $\endgroup$ – drhab Aug 14 '18 at 10:03
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    $\begingroup$ @drhab: Did you read the Wikipedia page I linked to? Hint: The interior of a Jordan curve is disjoint with the Jordan curve itself. $\endgroup$ – celtschk Aug 14 '18 at 10:05
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    $\begingroup$ @celtschk Admit that I did not read it carefully and now I see that you have a point. Nevertheless my conviction is that "disk" is in this context more appropriate than "circle". Maybe the OP can tell us what sort of interior he/she had in mind. $\endgroup$ – drhab Aug 14 '18 at 10:11
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Denote by $R$ the absolute value of a uniformly distributed random point $Z$ in the unit disc. The cdf and the pdf of $R$ are then given by $$F_1(s)=\left\{\eqalign{0\quad&(s<0) \cr s^2\quad&(0\leq s\leq1)\cr 1\quad&(s\geq1)\cr}\right.\ ,\qquad f_R(s)=\left\{\eqalign{0\quad&(s<0) \cr 2s\quad&(0< s< 1)\cr 0\quad&(s>1)\cr}\right.\ .\tag{1}$$ The index ${}_1$ refers to the first random point. Using $(1)$ one computes the cdf $F_2(s)$ of $R_1+R_2$ for two random points, and then $F_3(s)$, $F_4(s)$ as follows: $$F_2(s)=\int_0^1 f_R(t) F_1(s-t)\>dt,\quad\ldots, \quad F_4(s)=\int_0^1 f_R(t)F_3(s-t)\>dt\ .$$ Mathematica obtained the following expression for $F_4(s)$ in the interval $3<s<4$: $${31192 - 12288 s - 17920 s^2 + 12544 s^3 - 1680 s^4 - 448 s^5 + 112 s^6 - s^8\over2520}\ .$$ The probability $p$ that $R_1+R_2+R_3+R_4\geq\pi$ then comes to $$p=1-F_4(\pi)\approx0.1625\ .$$

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Guide:

  • Find the distribution of $R=\sqrt{X^2+Y^2}$ where $(X,Y)$ is chosen uniformly on the disk $\{(x,y)\in\mathbb R^2\mid x^2+y^2<1\}$. It is handsome to go for the CDF first. A PDF can be found by differentiating.

  • Find $\Pr(R_1+R_2+R_3+R_4>\pi)$ where the $R_i$ are independent and have the same distribution as $R$.

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Assume the disk have radius $1$.

If you have $r \sim Uni[0, 1]$, i.e. random radius and $\theta \sim Uni[0, 2\pi]$, i.e. random angle, then $X = \sqrt{r} \cos \theta$ and $Y = \sqrt{r} \sin \theta$ have a joint pdf on the disk $\frac{1}{\pi}$, i.e. describes a point picked uniformly at random on the disk.

And $R = \sqrt{X^2 + Y^2}$.

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