## Kainui one year ago Here's a fun problem I came across. During an hour two independent events can happen at any time. What's the probability that the events are at least 10 minutes apart?

1. rational

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2. Kainui

Hahaha yeah you got it.

3. rational

P(|E1-E2| > 10 minutes) = (Area of shaded region) / (Total area) = 5^2/6^2 = 25/36 ?

4. Kainui

Yep, exactly. =)

5. rational

these are really fun xD i got some practice a couple of months ago when @amistre64 was preparing for some probability theory exam

6. Kainui

Something that just occurred to me that I don't know the answer to is what is the probability of having 3 events 10 minutes apart? Or, is there a good reason why the probability is 5^2/6^2? If the events were separated by 20 minutes would the answer then become 4^2/6^2 ? I guess this would be easy to check for the general answer.

7. rational

very interesting, 3 events requires a triple integral is it

8. rational

second part of the question is easy yeah we always get squares in top and bottom because of symmetry : |E1-E2| > 20

9. Kainui

for the 2D case in general the answer is: $\frac{(60-t)^2}{60^2}$ which is quite nice!

10. ParthKohli

Wow, that's beautiful.

11. ParthKohli

Looks like my book has this... is this what is called "infinitistic probability"?

12. Kainui

Yup, I guess? I guess continuous probability distribution maybe? It's important for quantum mechanics!

13. ikram002p

:)

14. dan815

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15. dan815

oh greater so 5/6

16. dan815

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17. dan815

is that it?

18. dan815

i see that for one case its, 5/6 then therefore for every other point its always 5/6 of the total there so 5/6 all the time :)

19. dan815

i solved it by looking at a number line

20. dan815

|dw:1433020285038:dw|

21. dan815

so i thought about summing all these individual events up (kind of like an integral) however its constant here its always 5/6 of the smaller case, that event 1 happens at the one single point

22. dan815

Hence 5/6 for the complete thing

23. dan815

here is also something interesting... prolly not so much but kinda cool to visualze lol

24. dan815

cause its not good to think ofa number line but a circular number like for this or like a clock

25. dan815

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26. dan815

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27. Kainui

the problem is your answer seems to be wrong, the answer is 25/36 not 5/6 for the first question

28. dan815

ya i got that i change it xD

29. dan815

for a sec i thought u were asking less than 10

30. dan815

oh dang really

31. dan815

why is it 5^/6^2?

32. dan815

lemmee think

33. Kainui

this picture: |dw:1433020730099:dw| imagine each point is when two events happen, so all the points on the diagonal happen simultaneously. so for instance when one event happens, (10,20) at that point, that will fall in the black, so it's greater or equal to 10 min right? So the ratio of the black region to the entire square is the probability.

34. dan815

oh :O i get it

35. dan815

theres a problem here with assuming the circular thing too, because once event 1 occurs later

36. dan815

the prob is greater for 10 mins

37. dan815

:D

38. dan815

or wait no, the even can happen back in time too

39. Kainui

This is pretty similar to the Buffon needle problem which is a cool way to approximate $$\pi$$.

40. dan815

i dont think i understand the graph

41. Kainui

Honestly I don't think it gets much better than this general case: $\frac{(60-t)^2}{60^2}$ this tells you the probability that two random events happen with at least a separation of time t within an hour. Like if t=10 in the original then we get: $\frac{(60-10)^2}{60^2} = \frac{25}{36}$ and of course the probability that they're 0 minutes apart will give us 100% and probability they are 60 min apart will give us 0% just like we expect, it's just from calculating it based off of a variable triangle size inside cut out basically lol.

42. Kainui

|dw:1433021207310:dw| Here are some random points picked in there. The first represents some time that one event happened and the other represents the time the second event happened. So look along the diagonal, (0,0) means both events happened at hte beginning of the our. (30,30) means both events happened at the 30 minute mark. Makes sense? Now look at the point (10,50) that means the first event happened at 10 minutes and the other event happened at 50 minutes. This is one of the points that falls within "both events happening 10 minutes apart or greater" since these are obviously separated by 40 minutes yeah?

43. dan815

ah i gotcha now xD

44. dan815

the graph labelling didnt quite make sense to me, now its clear thanks

45. Kainui

cool so what if we have 3 events separated by 10 minutes, what does that cube look like?

46. dan815

|dw:1433021555908:dw|

47. Kainui

You would think so, that was my first guess too, but it doesn't seem to be right if you think of evaluating say, this point: |dw:1433021616607:dw|

48. dan815

ok lemme see more algrebrically first a+b+c<60*3 and diff between a,b,c is greater than 10

49. dan815

(5/6)^3?

50. dan815

^ thats a guess

51. dan815

wait what is the question?is it all 3 have to be apart by 10? or as long as all 3 dont happen in 10

52. Kainui

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53. dan815

ok wat u drew there looks like its as long all 3 dont happen within 10, so u can have 2 with in 10 and other out

54. Kainui

So one example of an event can be like (10, 20, 30) since they are all separated by 10 min.

55. dan815

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56. Kainui

yeah what I drew is wrong I think idk haha

57. Kainui

Yeah I just didn't wanna draw it with those lines in there cause it looked tacky

58. dan815

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59. dan815

lol

60. dan815

oh u can think of the 2by 2 case as 1 big event again

61. dan815

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62. dan815

so u solve for cases, when event 1 and 2 are close, and when they are not

63. dan815

eh nvm i feel like there is something easier to compute off the differences

64. dan815

did u already solve it?

65. Kainui

Nahhhh

66. dan815

ok :) im on it then sir

67. Kainui

some useful things: $$20 \le x+y+z \le 40$$ $$10 \le x+y \le 50$$ $$10 \le x+z \le 50$$ $$10 \le y+z \le 50$$

68. dan815

that what i was lookign at

69. dan815

like can we write out system of equationa for lines

70. dan815

and break it down into 4 or 6 or 9 cases or something

71. Kainui

I think this is enough to make up an integral, those are our bounds

72. dan815

hey btw the first lines ganeshew drew was

73. dan815

y=x+10 and x=y+10 right

74. Kainui

yeah

75. Kainui

notice they're symmetric so we only needed to find one part of that, then multiply by 2

76. dan815

ugh its gonna into these inequality cases