Welcome to our community

Be a part of something great, join today!

Selecting three points on a circle

Pranav

Well-known member
Nov 4, 2013
428
(I am not sure if this is the right place to post it but since my solution involves some Calculus, I decided to post it here.)

(Also, I am not well versed with the correct words and terms to be used while doing geometrical probability problems, I am sorry if I write something wrong. :eek: )

Problem:
Three points are selected on circumference of a circle. Find the probability that they lie on a semicircle.

Attempt:


$\angle POX=\theta$
$\angle QOX=\alpha$
$\angle ROX=\beta$

I select three arcs on the circle of length $Rd\theta$, $Rd\alpha$ and $Rd\beta$. The probability for P,Q and R to lie in these arc lengths is $\frac{d\theta}{2\pi}$, $\frac{d\alpha}{2\pi}$ and $\frac{d\beta}{2\pi}$ respectively. Hence,

$$dP=\frac{1}{8\pi^3}\,d\theta\,d\alpha\, d\beta$$

The limits for $\beta$ are from $-\pi+\alpha$ to $\pi+\theta$ but these limits only work when $\alpha$ lies from $\theta$ to $\pi+\theta$ and $\theta$ lies from 0 to $2\pi$. Solving the definite integral with these limits and multiplying the result by two gives the correct answer.

My question is whether there is way to identify the correct limits for $\alpha$ and $\beta$ in the first try. I spent quite a lot of time to figure out the correct limits. I initially used the limits $-\pi+\alpha$ to $\pi+\theta$ for $\beta$ and 0 to $2\pi$ for $\alpha$ which gives the incorrect answer. I tried checking different cases to see if my initial limits holds and then I arrived at the correct limits which took up a lot of time. Is there a nicer way to figure out the limits?

Any help is appreciated. Thanks!

(I have never formally studied triple integrals, the above integral was set up completely on my intuition so I am not sure if I have done it correctly.)
 
Last edited by a moderator:

chisigma

Well-known member
Feb 13, 2012
1,704
(I am not sure if this is the right place to post it but since my solution involves some Calculus, I decided to post it here.)

(Also, I am not well versed with the correct words and terms to be used while doing geometrical probability problems, I am sorry if I write something wrong. :eek: )

Problem:
Three points are selected on circumference of a circle. Find the probability that they lie on a semicircle.

Attempt:


$\angle POX=\theta$
$\angle QOX=\alpha$
$\angle ROX=\beta$

I select three arcs on the circle of length $Rd\theta$, $Rd\alpha$ and $Rd\beta$. The probability for P,Q and R to lie in these arc lengths is $\frac{d\theta}{2\pi}$, $\frac{d\alpha}{2\pi}$ and $\frac{d\beta}{2\pi}$ respectively. Hence,

$$dP=\frac{1}{8\pi^3}\,d\theta\,d\alpha\, d\beta$$

The limits for $\beta$ are from $-\pi+\alpha$ to $\pi+\theta$ but these limits only work when $\alpha$ lies from $\theta$ to $\pi+\theta$ and $\theta$ lies from 0 to $2\pi$. Solving the definite integral with these limits and multiplying the result by two gives the correct answer.

My question is whether there is way to identify the correct limits for $\alpha$ and $\beta$ in the first try. I spent quite a lot of time to figure out the correct limits. I initially used the limits $-\pi+\alpha$ to $\pi+\theta$ for $\beta$ and 0 to $2\pi$ for $\alpha$ which gives the incorrect answer. I tried checking different cases to see if my initial limits holds and then I arrived at the correct limits which took up a lot of time. Is there a nicer way to figure out the limits?

Any help is appreciated. Thanks!

(I have never formally studied triple integrals, the above integral was set up completely on my intuition so I am not sure if I have done it correctly.)
http://mathhelpboards.com/questions...stics-questions-other-sites-932.html#post4878

Kind regards

$\chi$ $\sigma$
 

Pranav

Well-known member
Nov 4, 2013
428

chisigma

Well-known member
Feb 13, 2012
1,704
More precisely the probability that, being the fist point in (1,0), the other two points lie in the up half circle is...


$\displaystyle P = \frac{1}{4}\ \int_{0}^{1} d x \int_{x}^{1} d y = \frac{1}{4}\ \int_{0}^{1} (1 - x)\ d x = \frac{1}{8}\ (1)$

... and the same if the other two points lie in the down half circle, so that the total probability is $\frac{1}{4}$...


Kind regards


$\chi$ $\sigma$
 

Pranav

Well-known member
Nov 4, 2013
428
More precisely the probability that, being the fist point in (1,0), the other two points lie in the up half circle is...


$\displaystyle P = \frac{1}{4}\ \int_{0}^{1} d x \int_{x}^{1} d y = \frac{1}{4}\ \int_{0}^{1} (1 - x)\ d x = \frac{1}{8}\ (1)$

... and the same if the other two points lie in the down half circle, so that the total probability is $\frac{1}{4}$...


Kind regards


$\chi$ $\sigma$
I am not sure if I still get it but is their really no need to define three angles? I mean, can we fix one of the points and deal with only $\alpha$ and $\beta$? :confused:

I am confused by the definite integral you set up. Why not integrate over the complete circle? I know it gives the incorrect answer but I don't see why you should divide the circle into two separate semicircles and then do the integration. :confused:
 
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
I apologize for the fact that my answer can have create some misunderstanding but in fact i remembered that o problem similar was resolved more that one year ago in the section 'unsolved statistic problems from other sites'. In that post it was required to compute the probability that, given three random points A, B and C on a circl, the center of the circle lies inside the triangle ABC, and the computed probability was found to be $P= \frac{1}{4}$. In Pranav's question it is required, given three random points A,B and C, to computed the probability that the three points lie on a half circle, which is equivalent to say that the center of the circle lies outside the triangle ABC, so that the requested probability is $P = 1 - \frac{1}{4} = \frac{3}{4}$...

Kind regards

$\chi$ $\sigma$
 

Pranav

Well-known member
Nov 4, 2013
428
I apologize for the fact that my answer can have create some misunderstanding but in fact i remembered that o problem similar was resolved more that one year ago in the section 'unsolved statistic problems from other sites'. In that post it was required to compute the probability that, given three random points A, B and C on a circl, the center of the circle lies inside the triangle ABC, and the computed probability was found to be $P= \frac{1}{4}$. In Pranav's question it is required, given three random points A,B and C, to computed the probability that the three points lie on a half circle, which is equivalent to say that the center of the circle lies outside the triangle ABC, so that the requested probability is $P = 1 - \frac{1}{4} = \frac{3}{4}$...

Kind regards

$\chi$ $\sigma$
Hi chisigma!

Both the problems are very similar and both seem to take the same approach. My question is that why we need to separate the circles into two semicircles?

Let get back to my original question. The limit for $\beta$ is from $-\pi+\alpha$ to $\pi+\theta$. As I said before, these limits work only when $\alpha$ lies from $\theta$ to $\pi+\theta$. This is equivalent to dividing the circles into two semicircles. Now if I make $\alpha$ vary from 0 to $2\pi$, I get an incorrect answer i.e I did not divide the circle into separate semicircles. Why do I get an incorrect answer? I mean, where is the error? I know the limits for $\beta$ I used doesn't work for the other half of semicircle but I couldn't figure out the error for quite some time and you don't really have that amount of time in the exams. So I want to know if there is a way to figure out the correct limits at the first sight.

Thanks!