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Problem Statement: Three dots are randomly placed in a circle of radius one cm. What is the probability that when a fourth dot is placed (randomly) in the circle, it is at least one cm away any of the other three dots?

Other Things I Should Probably Mention: I came up with this problem, and you don't need to evaulate your final expression since it's kinda ugly.

My Solution: WLOG let the fourth point be placed first. WLOG we only need to consider the probability when the fourth point is placed on a specific radii since the probability is the same for all possible radii of the circle (this part is kind of tricky to word, we basically choose a radii and only place the fourth point [randomly] on it, instead of considering the entire circle). Let $p(x)$ be the probability where $x$ represents the distance from our fourth point to the center of our circle. The function will be one minus the intersection of the areas of our circle and a circle of radius one centered at the fourth point, divided by the area of our circle ($\pi$), and then the whole thing cubed since there are $3$ points in total. Finding the intersection can be done using trigonometry or calculus, I used calculus and got $$I(x)=2\left(\frac{\pi}{2}-\left(\frac{x}{2}\sqrt{1-\frac{x^2}{4}}+\arcsin\left(\frac{x}{2}\right)\right)\right)$$ Thus, $$p(x) = \left(1-\frac{I(x)}{\pi}\right)^3$$ Now to find the probability, we integrate $p(x)$ from $0$ to $1$ to cover all the possible values of $x$. Thus the answer is $$ \int_0^1 \left( 1-\left(1-\left(\frac{x}{\pi}\sqrt{1-\frac{x^2}{4}}+\frac{2}{\pi}\arcsin\left(\frac{x}{2}\right)\right)\right) \right)^3\,dx $$ $$ \int_0^1 \left(\frac{x}{\pi}\sqrt{1-\frac{x^2}{4}}+\frac{2}{\pi}\arcsin\left(\frac{x}{2}\right)\right)^3\,dx $$ Numerically, this is approximately $\approx 0.059$.

My question is if everything I did was okay since I'm not very experienced working with probability density functions. Also since there are no official solutions (since I came up with the problem), so I'm curious to see if my answer is correct or not.

Update: Thanks to @achillehui for pointing out $x$ is not uniformly distributed by the nature of the circle, in fact its' PDF is $2x$. Thus the integral should be $$ \int_0^1 \left(\frac{x}{\pi}\sqrt{1-\frac{x^2}{4}}+\frac{2}{\pi}\arcsin\left(\frac{x}{2}\right)\right)^32x\,dx $$ Which gives the correct answer of around 0.094015.

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    $\begingroup$ If the fourth point is chosen uniformly from the unit disk, then $x$ is not uniformly distributed over $[0,1]$. Instead, its PDF is $2x$. The probability should be $$\int_0^1 \left(\frac{x}{\pi}\sqrt{1-\frac{x^2}{4}}+\frac{2}{\pi}\arcsin\left(\frac{x}{2}\right)\right)^3 \color{red}{2x}\,dx \approx 0.094015$$ $\endgroup$ – achille hui Dec 15 '18 at 5:13
  • $\begingroup$ @achillehui Thanks for the response! What you said makes sense but 0.094 seems too high for the probability. I wrote a python script to approximate the answer and after several million cycles the probability hovers around ~0.0500. So I think there might be something else as well, I'll update if I find anything. $\endgroup$ – biscuit Dec 15 '18 at 17:05
  • $\begingroup$ Don't approximate your answer and don't do any shortcut. Go directly to original problem by sampling 4 random points from the unit disk and count the number of times the first point is at a distance $\ge 1$ from all 3 other points. I get $94115$ out a $10^6$ simulation. $\endgroup$ – achille hui Dec 15 '18 at 17:51
  • $\begingroup$ Yep it appears you're right! I had an error in my programming which caused it to pick skewed random points. It appears that the answer now agrees with my simulation. Thanks a lot! $\endgroup$ – biscuit Dec 15 '18 at 19:01

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