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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
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|dw:1351969912033:dw| Something like this.
yes, thats the case when centre does not line inside circle...
*lie

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Other answers:

and i need probability
I wonder if coordinate geometry could be helpful here? To get a name on the center point and the two random points...
i tried taking centre as origin and using distance formula....but its getting too complicated and no hopes of finding probability
*centre of square
What is the condition that the point is in the circle? In the coordinates? If you found it, please write.
|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 ?
It is a very good condition! Try to use it in polar system.
Two points with coordinates \((\rho_1,\varphi_1)\) and \((\rho_2,\varphi_2)\).
my brain suddenly goes into hibernate mode when it comes to polar co-ordinates :P and remember we have to find probability.
|dw:1351970854925:dw|
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?
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...
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.
u lost me here....
how square of side 2pi ?
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..
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?
You have only to plot this condition in the square. Please, say something.
|dw:1351972589720:dw|
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|
filled area = 9pi^2/ 8 ?
filled area = \((2\pi-\frac{\pi}2)^2\)
oh, yes....multipled by 2, twice, instead of once....so 9pi^2/16
sorry, 9pi^2/4
Now it is OK.
Now divide it by the area of the whole square and get final probability.
thats 9/16...but i am trying to understand that figure.....
i still don't understand why is the side of square =2pi
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|
randomly with uniform distribution ?
and no.
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.
So what with sniper?
2 quares of side 2R and R ?
Yes. They are concentric circles with radiuses R and 2R.
1/2 right?
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.
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.. :)
@AccessDenied That's right. Did you get the very first question? About 2 points?
I was not able to get it. I was trying my own method but couldn't figure it out. :(

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