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hartnn

  • 2 years ago

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.

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  1. klimenkov
    • 2 years ago
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    |dw:1351969912033:dw| Something like this.

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

  3. hartnn
    • 2 years ago
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    *lie

  4. hartnn
    • 2 years ago
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    and i need probability

  5. AccessDenied
    • 2 years ago
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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...

  6. hartnn
    • 2 years ago
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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

  7. hartnn
    • 2 years ago
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    *centre of square

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

  9. hartnn
    • 2 years ago
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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 ?

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

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

  12. hartnn
    • 2 years ago
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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.

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

  14. klimenkov
    • 2 years ago
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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?

  15. hartnn
    • 2 years ago
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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...

  16. klimenkov
    • 2 years ago
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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.

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

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

  19. klimenkov
    • 2 years ago
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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..

  20. klimenkov
    • 2 years ago
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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?

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

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

  23. klimenkov
    • 2 years ago
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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|

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

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

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

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

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

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

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

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

  32. klimenkov
    • 2 years ago
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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|

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

  34. hartnn
    • 2 years ago
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    and no.

  35. klimenkov
    • 2 years ago
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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.

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

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

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

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

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

  41. klimenkov
    • 2 years ago
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    @geerky42 No.

  42. hartnn
    • 2 years ago
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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.

  43. AccessDenied
    • 2 years ago
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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.. :)

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

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

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