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?

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

Beginner2022-10-03Added 7 answers

Step 1

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

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}\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

Beginner2022-10-04Added 1 answers

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

$\begin{array}{}\text{(1)}& {F}_{1}(s)=\{\begin{array}{rl}0\phantom{\rule{1em}{0ex}}& (s<0)\\ {s}^{2}\phantom{\rule{1em}{0ex}}& (0\le s\le 1)\\ 1\phantom{\rule{1em}{0ex}}& (s\ge 1)\end{array}\text{},\phantom{\rule{2em}{0ex}}{f}_{R}(s)=\{\begin{array}{rl}0\phantom{\rule{1em}{0ex}}& (s0)\\ 2s\phantom{\rule{1em}{0ex}}& (0s1)\\ 0\phantom{\rule{1em}{0ex}}& (s1)\end{array}\text{}.\end{array}$

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:

${F}_{2}(s)={\int}_{0}^{1}{f}_{R}(t){F}_{1}(s-t)\phantom{\rule{mediummathspace}{0ex}}dt,\phantom{\rule{1em}{0ex}}\dots ,\phantom{\rule{1em}{0ex}}{F}_{4}(s)={\int}_{0}^{1}{f}_{R}(t){F}_{3}(s-t)\phantom{\rule{mediummathspace}{0ex}}dt\text{}.$

Step 2

Mathematica obtained the following expression for ${F}_{4}(s)$ in the interval $3<s<4$:

$\frac{31192-12288s-17920{s}^{2}+12544{s}^{3}-1680{s}^{4}-448{s}^{5}+112{s}^{6}-{s}^{8}}{2520}\text{}.$

The probability p that ${R}_{1}+{R}_{2}+{R}_{3}+{R}_{4}\ge \pi $ then comes to $p=1-{F}_{4}(\pi )\approx 0.1625\text{}.$

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

$\begin{array}{}\text{(1)}& {F}_{1}(s)=\{\begin{array}{rl}0\phantom{\rule{1em}{0ex}}& (s<0)\\ {s}^{2}\phantom{\rule{1em}{0ex}}& (0\le s\le 1)\\ 1\phantom{\rule{1em}{0ex}}& (s\ge 1)\end{array}\text{},\phantom{\rule{2em}{0ex}}{f}_{R}(s)=\{\begin{array}{rl}0\phantom{\rule{1em}{0ex}}& (s0)\\ 2s\phantom{\rule{1em}{0ex}}& (0s1)\\ 0\phantom{\rule{1em}{0ex}}& (s1)\end{array}\text{}.\end{array}$

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:

${F}_{2}(s)={\int}_{0}^{1}{f}_{R}(t){F}_{1}(s-t)\phantom{\rule{mediummathspace}{0ex}}dt,\phantom{\rule{1em}{0ex}}\dots ,\phantom{\rule{1em}{0ex}}{F}_{4}(s)={\int}_{0}^{1}{f}_{R}(t){F}_{3}(s-t)\phantom{\rule{mediummathspace}{0ex}}dt\text{}.$

Step 2

Mathematica obtained the following expression for ${F}_{4}(s)$ in the interval $3<s<4$:

$\frac{31192-12288s-17920{s}^{2}+12544{s}^{3}-1680{s}^{4}-448{s}^{5}+112{s}^{6}-{s}^{8}}{2520}\text{}.$

The probability p that ${R}_{1}+{R}_{2}+{R}_{3}+{R}_{4}\ge \pi $ then comes to $p=1-{F}_{4}(\pi )\approx 0.1625\text{}.$

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