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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.

  • 2 years ago
  • 2 years ago

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  1. klimenkov Group Title
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    |dw:1351969912033:dw| Something like this.

    • 2 years ago
  2. hartnn Group Title
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    yes, thats the case when centre does not line inside circle...

    • 2 years ago
  3. hartnn Group Title
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    *lie

    • 2 years ago
  4. hartnn Group Title
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    and i need probability

    • 2 years ago
  5. AccessDenied Group Title
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    I wonder if coordinate geometry could be helpful here? To get a name on the center point and the two random points...

    • 2 years ago
  6. hartnn Group Title
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    i tried taking centre as origin and using distance formula....but its getting too complicated and no hopes of finding probability

    • 2 years ago
  7. hartnn Group Title
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    *centre of square

    • 2 years ago
  8. klimenkov Group Title
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    What is the condition that the point is in the circle? In the coordinates? If you found it, please write.

    • 2 years ago
  9. hartnn Group Title
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    |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 ?

    • 2 years ago
  10. klimenkov Group Title
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    It is a very good condition! Try to use it in polar system.

    • 2 years ago
  11. klimenkov Group Title
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    Two points with coordinates \((\rho_1,\varphi_1)\) and \((\rho_2,\varphi_2)\).

    • 2 years ago
  12. hartnn Group Title
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    my brain suddenly goes into hibernate mode when it comes to polar co-ordinates :P and remember we have to find probability.

    • 2 years ago
  13. klimenkov Group Title
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    |dw:1351970854925:dw|

    • 2 years ago
  14. klimenkov Group Title
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    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?

    • 2 years ago
  15. hartnn Group Title
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    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...

    • 2 years ago
  16. klimenkov Group Title
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    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.

    • 2 years ago
  17. hartnn Group Title
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    u lost me here....

    • 2 years ago
  18. hartnn Group Title
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    how square of side 2pi ?

    • 2 years ago
  19. klimenkov Group Title
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    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..

    • 2 years ago
  20. klimenkov Group Title
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    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?

    • 2 years ago
  21. klimenkov Group Title
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    You have only to plot this condition in the square. Please, say something.

    • 2 years ago
  22. hartnn Group Title
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    |dw:1351972589720:dw|

    • 2 years ago
  23. klimenkov Group Title
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    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|

    • 2 years ago
  24. hartnn Group Title
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    filled area = 9pi^2/ 8 ?

    • 2 years ago
  25. klimenkov Group Title
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    filled area = \((2\pi-\frac{\pi}2)^2\)

    • 2 years ago
  26. hartnn Group Title
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    oh, yes....multipled by 2, twice, instead of once....so 9pi^2/16

    • 2 years ago
  27. hartnn Group Title
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    sorry, 9pi^2/4

    • 2 years ago
  28. klimenkov Group Title
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    Now it is OK.

    • 2 years ago
  29. klimenkov Group Title
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    Now divide it by the area of the whole square and get final probability.

    • 2 years ago
  30. hartnn Group Title
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    thats 9/16...but i am trying to understand that figure.....

    • 2 years ago
  31. hartnn Group Title
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    i still don't understand why is the side of square =2pi

    • 2 years ago
  32. klimenkov Group Title
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    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|

    • 2 years ago
  33. hartnn Group Title
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    randomly with uniform distribution ?

    • 2 years ago
  34. hartnn Group Title
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    and no.

    • 2 years ago
  35. klimenkov Group Title
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    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.

    • 2 years ago
  36. klimenkov Group Title
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    So what with sniper?

    • 2 years ago
  37. hartnn Group Title
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    2 quares of side 2R and R ?

    • 2 years ago
  38. klimenkov Group Title
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    Yes. They are concentric circles with radiuses R and 2R.

    • 2 years ago
  39. klimenkov Group Title
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    @CalebBeavers @AccessDenied @geerky42 Anyone got it?

    • 2 years ago
  40. geerky42 Group Title
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    1/2 right?

    • 2 years ago
  41. klimenkov Group Title
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    @geerky42 No.

    • 2 years ago
  42. hartnn Group Title
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    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.

    • 2 years ago
  43. AccessDenied Group Title
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    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.. :)

    • 2 years ago
  44. klimenkov Group Title
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    @AccessDenied That's right. Did you get the very first question? About 2 points?

    • 2 years ago
  45. AccessDenied Group Title
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    I was not able to get it. I was trying my own method but couldn't figure it out. :(

    • 2 years ago
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