Tatiana Cook

2022-10-02

Finding Geometric Probability?
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?

Allvin03

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

Jamarcus Lindsey

Step 1
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

The index 1 refers to the first random point. Using (1) one computes the cdf ${F}_{2}\left(s\right)$ of ${R}_{1}+{R}_{2}$ for two random points, and then ${F}_{3}\left(s\right)$, ${F}_{4}\left(s\right)$ as follows:

Step 2
Mathematica obtained the following expression for ${F}_{4}\left(s\right)$ in the interval $3:

The probability p that ${R}_{1}+{R}_{2}+{R}_{3}+{R}_{4}\ge \pi$ then comes to

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