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hartnn
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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.
 one year ago
 one year ago
hartnn Group Title
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.
 one year ago
 one year ago

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klimenkov Group TitleBest ResponseYou've already chosen the best response.2
dw:1351969912033:dw Something like this.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
yes, thats the case when centre does not line inside circle...
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
and i need probability
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
I wonder if coordinate geometry could be helpful here? To get a name on the center point and the two random points...
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
i tried taking centre as origin and using distance formula....but its getting too complicated and no hopes of finding probability
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
*centre of square
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
What is the condition that the point is in the circle? In the coordinates? If you found it, please write.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
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 ?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
It is a very good condition! Try to use it in polar system.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Two points with coordinates \((\rho_1,\varphi_1)\) and \((\rho_2,\varphi_2)\).
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
my brain suddenly goes into hibernate mode when it comes to polar coordinates :P and remember we have to find probability.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
dw:1351970854925:dw
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
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...
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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:dwAny 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.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
u lost me here....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
how square of side 2pi ?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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..
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
You have only to plot this condition in the square. Please, say something.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
dw:1351972589720:dw
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
filled area = 9pi^2/ 8 ?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
filled area = \((2\pi\frac{\pi}2)^2\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
oh, yes....multipled by 2, twice, instead of once....so 9pi^2/16
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
sorry, 9pi^2/4
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Now it is OK.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Now divide it by the area of the whole square and get final probability.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
thats 9/16...but i am trying to understand that figure.....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
i still don't understand why is the side of square =2pi
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
randomly with uniform distribution ?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
So what with sniper?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
2 quares of side 2R and R ?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
Yes. They are concentric circles with radiuses R and 2R.
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
@CalebBeavers @AccessDenied @geerky42 Anyone got it?
 one year ago

geerky42 Group TitleBest ResponseYou've already chosen the best response.0
1/2 right?
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
@geerky42 No.
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.3
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.
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
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.. :)
 one year ago

klimenkov Group TitleBest ResponseYou've already chosen the best response.2
@AccessDenied That's right. Did you get the very first question? About 2 points?
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.0
I was not able to get it. I was trying my own method but couldn't figure it out. :(
 one year ago
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