hartnn
  • hartnn
The two points are uniformly and independently marked inside a square . What is the probability that the center of the square is inside the circle formed by the two points as the diameter.
Mathematics
jamiebookeater
  • jamiebookeater
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klimenkov
  • klimenkov
|dw:1351969912033:dw| Something like this.
hartnn
  • hartnn
yes, thats the case when centre does not line inside circle...
hartnn
  • hartnn
*lie

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hartnn
  • hartnn
and i need probability
AccessDenied
  • AccessDenied
I wonder if coordinate geometry could be helpful here? To get a name on the center point and the two random points...
hartnn
  • hartnn
i tried taking centre as origin and using distance formula....but its getting too complicated and no hopes of finding probability
hartnn
  • hartnn
*centre of square
klimenkov
  • klimenkov
What is the condition that the point is in the circle? In the coordinates? If you found it, please write.
hartnn
  • hartnn
|dw:1351970387637:dw| the centre , if it lies on the circle(or inside it) will form 90 degree angle (or more) as shown.... but this doen't help, i think.....or does it ?
klimenkov
  • klimenkov
It is a very good condition! Try to use it in polar system.
klimenkov
  • klimenkov
Two points with coordinates \((\rho_1,\varphi_1)\) and \((\rho_2,\varphi_2)\).
hartnn
  • hartnn
my brain suddenly goes into hibernate mode when it comes to polar co-ordinates :P and remember we have to find probability.
klimenkov
  • klimenkov
|dw:1351970854925:dw|
klimenkov
  • klimenkov
If you have 2 points with polar coordinates (ρ1,φ1) and (ρ2,φ2), your condition will reduce the number of variables. \(|\varphi_1-\varphi_2|\ge\frac{\pi}2\). Do you get it?
hartnn
  • hartnn
that i got... will that help in finding the probability ? also we still haven't used the info. that points are 'uniformly' marked....that is their pdf is of uniform distribution...
klimenkov
  • klimenkov
Now try to get. \(\varphi_1\) and \(\varphi_2\) varies in \([0,2\pi]\). So, we can draw this as a square with a length equal to 2 pi:|dw:1351971559152:dw|Any point of this square will represent 2 points. One question is "What about \(\rho_1,\rho_2\)?" Actually the answer is that the probability depends only on \(\varphi\) - the angle and not on the distance. So, you just have to plot |φ1−φ2|≥π/2 in the square on the picture. Please do it.
hartnn
  • hartnn
u lost me here....
hartnn
  • hartnn
how square of side 2pi ?
klimenkov
  • klimenkov
We got a condition that the origin lies in the circle, right? It is \(|\varphi_1-\varphi_2|\ge\frac{\pi}2\). That's ok? May be my English should be better..
klimenkov
  • klimenkov
Now we use that fact that the points are uniformly marked. Randomly. So for a point it can be any angle from \([0,2\pi]\). So, it will be more demonstrably if we will plot values of this two angles is this way. It will be easier later to find probability. The angles of any point represent the point of the square. So we can say that any point of the square gives two angles of our random points. Got it?
klimenkov
  • klimenkov
You have only to plot this condition in the square. Please, say something.
hartnn
  • hartnn
|dw:1351972589720:dw|
klimenkov
  • klimenkov
No, it will be like this. Just find the filled area and divide it by \(4\pi^2\) - the area of the square.|dw:1351973075885:dw|
hartnn
  • hartnn
filled area = 9pi^2/ 8 ?
klimenkov
  • klimenkov
filled area = \((2\pi-\frac{\pi}2)^2\)
hartnn
  • hartnn
oh, yes....multipled by 2, twice, instead of once....so 9pi^2/16
hartnn
  • hartnn
sorry, 9pi^2/4
klimenkov
  • klimenkov
Now it is OK.
klimenkov
  • klimenkov
Now divide it by the area of the whole square and get final probability.
hartnn
  • hartnn
thats 9/16...but i am trying to understand that figure.....
hartnn
  • hartnn
i still don't understand why is the side of square =2pi
klimenkov
  • klimenkov
Can you solve this problem? The sniper hits randomly at the big circle. Any of his shot is in this circle. What is the probability to hit the small circle?|dw:1351973773130:dw|
hartnn
  • hartnn
randomly with uniform distribution ?
hartnn
  • hartnn
and no.
klimenkov
  • klimenkov
It is because the angle of the point in polar coordinates varies from \(0\) to \(2\pi\). So any of this values can be the angle of the point in polar coordinates.
klimenkov
  • klimenkov
So what with sniper?
hartnn
  • hartnn
2 quares of side 2R and R ?
klimenkov
  • klimenkov
Yes. They are concentric circles with radiuses R and 2R.
klimenkov
  • klimenkov
@CalebBeavers @AccessDenied @geerky42 Anyone got it?
geerky42
  • geerky42
1/2 right?
klimenkov
  • klimenkov
@geerky42 No.
hartnn
  • hartnn
how does a circle become a square? ...... nevermind, got it.....i will go through this again later, i can't keeps my eyes open now.
AccessDenied
  • AccessDenied
Prob of hitting smaller circle = Area of smaller circle / Area of bigger circle = (pi R^2)/(pi (2R)^2) = 1/4 That's how I'd approach it, anyways.. :)
klimenkov
  • klimenkov
@AccessDenied That's right. Did you get the very first question? About 2 points?
AccessDenied
  • AccessDenied
I was not able to get it. I was trying my own method but couldn't figure it out. :(

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