## 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

1. klimenkov Group Title

|dw:1351969912033:dw| Something like this.

2. hartnn Group Title

yes, thats the case when centre does not line inside circle...

3. hartnn Group Title

*lie

4. hartnn Group Title

and i need probability

5. AccessDenied Group Title

I wonder if coordinate geometry could be helpful here? To get a name on the center point and the two random points...

6. hartnn Group Title

i tried taking centre as origin and using distance formula....but its getting too complicated and no hopes of finding probability

7. hartnn Group Title

*centre of square

8. klimenkov Group Title

What is the condition that the point is in the circle? In the coordinates? If you found it, please write.

9. hartnn Group Title

|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 Group Title

It is a very good condition! Try to use it in polar system.

11. klimenkov Group Title

Two points with coordinates $$(\rho_1,\varphi_1)$$ and $$(\rho_2,\varphi_2)$$.

12. hartnn Group Title

my brain suddenly goes into hibernate mode when it comes to polar co-ordinates :P and remember we have to find probability.

13. klimenkov Group Title

|dw:1351970854925:dw|

14. klimenkov Group Title

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 Group Title

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 Group Title

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 Group Title

u lost me here....

18. hartnn Group Title

how square of side 2pi ?

19. klimenkov Group Title

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 Group Title

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 Group Title

You have only to plot this condition in the square. Please, say something.

22. hartnn Group Title

|dw:1351972589720:dw|

23. klimenkov Group Title

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 Group Title

filled area = 9pi^2/ 8 ?

25. klimenkov Group Title

filled area = $$(2\pi-\frac{\pi}2)^2$$

26. hartnn Group Title

oh, yes....multipled by 2, twice, instead of once....so 9pi^2/16

27. hartnn Group Title

sorry, 9pi^2/4

28. klimenkov Group Title

Now it is OK.

29. klimenkov Group Title

Now divide it by the area of the whole square and get final probability.

30. hartnn Group Title

thats 9/16...but i am trying to understand that figure.....

31. hartnn Group Title

i still don't understand why is the side of square =2pi

32. klimenkov Group Title

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 Group Title

randomly with uniform distribution ?

34. hartnn Group Title

and no.

35. klimenkov Group Title

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 Group Title

So what with sniper?

37. hartnn Group Title

2 quares of side 2R and R ?

38. klimenkov Group Title

Yes. They are concentric circles with radiuses R and 2R.

39. klimenkov Group Title

@CalebBeavers @AccessDenied @geerky42 Anyone got it?

40. geerky42 Group Title

1/2 right?

41. klimenkov Group Title

@geerky42 No.

42. hartnn Group Title

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 Group Title

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 Group Title

@AccessDenied That's right. Did you get the very first question? About 2 points?

45. AccessDenied Group Title

I was not able to get it. I was trying my own method but couldn't figure it out. :(