Point in a circle

A circle of radius 1 is randomly chosen strictly inside a circle of radius 2. A point is chosen, again at random, inside the circle with radius 1. What is the probability this point is closer to the center of the circle with radius 2 than the circumference of the circle with radius 2?

Note by Daniel Liu
5 years, 10 months ago

No vote yet
5 votes

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Edited to match the corrected question...

Let RR be the distance of the centre O2O_2 of the circle CC of radius 11 from the centre O1O_1 of the circle of radius 22. Since O2O_2 can be anywhere inside the circle of radius 11 with centre O1O_1 (shown dotted in the diagram) we have P[Rr]=πr2π=r2\mathbb{P}[R \le r] \,=\, \tfrac{\pi r^2}{\pi} \,=\, r^2 for any 0r10 \le r \le 1 and hence RR has the probability distribution function f(r)  =  {2r0r10o.w. f(r) \; = \; \left\{ \begin{array}{ll} 2r & 0 \le r \le 1 \\ 0 & \mbox{o.w.} \end{array} \right.

alternate<em>text alternatetext

Given that R=rR = r, the probability that the randomly chosen point in the circle CC is closer to the centre of the original circle than its circumference is the probability that the randomly chosen point lies inside the region common to CC and the (dotted) circle of centre O1O_1 and radius 11, which is P[b>aR=r]=2π(θsinθcosθ)=1π[2cos1(r2)r114r2] \begin{array}{rcl} \mathbb{P}[b > a | R = r] & = & \tfrac{2}{\pi}(\theta - \sin\theta\cos\theta) \\ & = & \tfrac{1}{\pi}\left[2\cos^{-1}(\tfrac{r}{2}) - r\sqrt{1 - \tfrac14r^2}\right] \end{array} and hence P[b>a]=01P[b>aR=r]2rdr  =  2π01(2rcos1(r2)r2114r2)dr=2π12π13π(4ucosu4cos2usinu)(2sinu)du=2π13π12π(4usin2u2sin22u)du=2π23ππ(vsinvsin2v)dv=2π[vcosv+sinv12v+14sin2v]23ππ=2π×18(4π33)  =  1334π \begin{array}{rcl} \mathbb{P}[b > a] & = & \displaystyle\int_0^1 \mathbb{P}[b > a|R=r]\,2r\,dr \; = \; \frac{2}{\pi}\int_0^1\left(2r\cos^{-1}(\tfrac{r}{2}) - r^2\sqrt{1 - \tfrac14r^2}\right)\,dr \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac12\pi}^{\frac13\pi}\left(4u\cos u - 4\cos^2u\sin u\right)(-2\sin u)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac13\pi}^{\frac12\pi}\left(4u\sin2u - 2\sin^22u\right)\,du \\ & = & \displaystyle\frac{2}{\pi}\int_{\frac23\pi}^\pi \left(v\sin v - \sin^2v\right)\,dv \\ & = & \displaystyle\frac{2}{\pi}\left[ -v\cos v + \sin v - \tfrac12v + \tfrac14\sin2v\right]_{\frac23\pi}^\pi \\ & = & \frac{2}{\pi} \times \frac{1}{8}(4\pi - 3\sqrt{3}) \; = \; 1 - \frac{3\sqrt{3}}{4\pi} \end{array}

Mark Hennings - 5 years, 10 months ago

Log in to reply

Sorry, little typo there. Fixed. That would mean the probability is 4π334π\dfrac{4\pi-3\sqrt{3}}{4\pi}, correct? I never thought this problem would actually be that hard.

Daniel Liu - 5 years, 10 months ago

Log in to reply

Hi,

I tried out a simulation, choosing the coordinates (r1cosθ+r2cosα,r1sinθ+r2sinα) (r_1\, \cos \theta+r_2\, \cos \alpha ,r_1\, \sin \theta+r_2\, \sin \alpha ) at random, where 0r1,r210\le r_1, r_2 \le 1 and 0θ,α2π0\le \theta, \alpha \le 2\,\pi.

The distance from the center of the bigger circle, a=r12+r22+2r1r2cos(θα) a=\sqrt{r_1^2+r_2^2+2\, r_1\, r_2\, \cos {(\theta-\alpha)}} and b=2ab = 2 -a, simulating for 1000000 trials, I got the average b>ab > a about 0.777, which is more than 0.5865.

Can't see any mistake, though.

gopinath no - 5 years, 10 months ago

Log in to reply

Did you assume that r1r_1 and r2r_2 were uniform on [0,1][0,1]?

Mark Hennings - 5 years, 10 months ago

Log in to reply

@Mark Hennings Yes, I did..

gopinath no - 5 years, 10 months ago

Log in to reply

@Gopinath No The pdf of r1r_1 and r2r_2 is 2x2x for 0x10\le x \le 1. Look at the beginning of my solution.

Mark Hennings - 5 years, 10 months ago

Log in to reply

@Mark Hennings I saw, but I thought it was required only for deriving the answer analytically.

r1r_1 is the distance from the center of the circle with radius 2 centered at (0,0)(0,0), and get the coordinates (r1cosθ,r1sinθ) (r_1\cos \theta, r_1 \sin \theta ), which is the center of the smaller circle. Then select (r2cosα,r2sinα) (r_2\cos \alpha, r_2 \sin \alpha ), which is the point within the smaller circle. Then I did my simulation as in the earlier post.

Is there something wrong with this thinking? Aren't the variables independent?

gopinath no - 5 years, 10 months ago

Log in to reply

@Gopinath No If you are going to simulate this, you must adjust your simulation so that the chances of choosing r1r_1 and r2r_2 correct. You can ignore θ\theta. Provided that you choose r1r_1 and r2r_2 independently from the correct pdf, and choose α\alpha randomly and uniformly in [0,2π)[0,2\pi), all should be well. It will not work if you choose r1r_1 and r2r_2 randomly from a uniform distribution.

To model r1r_1 and r2r_2 correctly, choose r12r_1^2 and r22r_2^2 from a uniform distribution on [0,1][0,1]. If you do that, your simulation should agree with my theoretical result (my one did).

Mark Hennings - 5 years, 10 months ago

Log in to reply

@Mark Hennings Oh, I see! It's agreeing now.

Thanks for your help, much appreciated.

gopinath no - 5 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...