Answer To A Probability Problem About Placing Random Points In A Circle.
$begingroup$
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 $$I(x)=2left(frac{pi}{2}-left(frac{x}{2}sqrt{1-frac{x^2}{4}}+arcsinleft(frac{x}{2}right)right)right)$$ 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 $$ int_0^1 left( 1-left(1-left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)right) right)^3,dx $$
$$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3,dx $$
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 $$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^32x,dx $$
Which gives the correct answer of around 0.094015.
calculus probability integration density-function geometric-probability
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
$endgroup$
add a comment |
$begingroup$
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 $$I(x)=2left(frac{pi}{2}-left(frac{x}{2}sqrt{1-frac{x^2}{4}}+arcsinleft(frac{x}{2}right)right)right)$$ 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 $$ int_0^1 left( 1-left(1-left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)right) right)^3,dx $$
$$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3,dx $$
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 $$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^32x,dx $$
Which gives the correct answer of around 0.094015.
calculus probability integration density-function geometric-probability
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
$endgroup$
1
$begingroup$
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 $$int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3 color{red}{2x},dx approx 0.094015$$
$endgroup$
– 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
add a comment |
$begingroup$
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 $$I(x)=2left(frac{pi}{2}-left(frac{x}{2}sqrt{1-frac{x^2}{4}}+arcsinleft(frac{x}{2}right)right)right)$$ 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 $$ int_0^1 left( 1-left(1-left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)right) right)^3,dx $$
$$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3,dx $$
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 $$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^32x,dx $$
Which gives the correct answer of around 0.094015.
calculus probability integration density-function geometric-probability
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
$endgroup$
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 $$I(x)=2left(frac{pi}{2}-left(frac{x}{2}sqrt{1-frac{x^2}{4}}+arcsinleft(frac{x}{2}right)right)right)$$ 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 $$ int_0^1 left( 1-left(1-left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)right) right)^3,dx $$
$$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3,dx $$
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 $$ int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^32x,dx $$
Which gives the correct answer of around 0.094015.
calculus probability integration density-function geometric-probability
calculus probability integration density-function geometric-probability
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
edited Dec 15 '18 at 19:07
biscuit
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
asked Dec 15 '18 at 3:52
biscuitbiscuit
62
62
1
$begingroup$
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 $$int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3 color{red}{2x},dx approx 0.094015$$
$endgroup$
– 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
add a comment |
1
$begingroup$
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 $$int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3 color{red}{2x},dx approx 0.094015$$
$endgroup$
– 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
1
1
$begingroup$
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 $$int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3 color{red}{2x},dx approx 0.094015$$
$endgroup$
– achille hui
Dec 15 '18 at 5:13
$begingroup$
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 $$int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3 color{red}{2x},dx approx 0.094015$$
$endgroup$
– 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
$begingroup$
@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
$begingroup$
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
$begingroup$
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
add a comment |
0
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1
$begingroup$
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 $$int_0^1 left(frac{x}{pi}sqrt{1-frac{x^2}{4}}+frac{2}{pi}arcsinleft(frac{x}{2}right)right)^3 color{red}{2x},dx approx 0.094015$$
$endgroup$
– 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