Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

klimenkov

  • 2 years ago

Two persons want to meet. They know the place where to meet. The time when one of them come to the place of meeting is equiprobable and lies in the \([0,T]\). Someone of them who comes first will wait \(\tau\) minutes and go away. What is the probability of meeting?

  • This Question is Closed
  1. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't seem to understand this problem ... are they allowed to come at any time?

  2. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes. Any time from 0 to T. For example you want to meet your friend between 1 p.m and 2 p.m. This is the same.

  3. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \([0,T]\) is used to simplify.

  4. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    and how long will I wait?

  5. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    For example, 15 minutes = 1/4 hour.

  6. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm not good with probability ... but this question seems interesting ... since most of my friends always like about time and distance while waiting.

  7. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    If you are interested I will give you a solution.

  8. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But firstly, I'd like you to solve another problem.

  9. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hold on ... I'll wait.

  10. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    If there are n balls in the very dark room and among them are m white and other n-m are black. What is the probability to take a white ball if you can see nothing and the probabilities for taking any of n balls are the same (equiprobable)?

  11. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    m/n??

  12. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes. Now lets try more complicated what is called a geometric probability. Someone want to hit the zone #1 by shooting from a gun. What is the probability to shot at zone #1 if the shooter always hit in the big circle? |dw:1350222646524:dw|

  13. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    pir^2/4pir^2 = 1/4 ??

  14. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes. Very good! So the probability is the ratio of the areas! The method for my problem is very similar.

  15. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    all right ... let's get down to this.|dw:1350223223318:dw|

  16. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1350223247938:dw|

  17. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hint! Let x is the time for the first person for example - for you. And y is for the second one - your friend. So the situation when you will come can be described as the pair \((x,y)\).

  18. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    No!!! The last one is wrong. Try to get why.

  19. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol ... first one can arrive at any time ...it wouldn't matter.

  20. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    They both can arrive at any time!

  21. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1350223462963:dw|

  22. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Better to say - you both :)

  23. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hm.. Can't get what you do...

  24. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Lets come back to the shooter. What is the probability to hit in the center of the circle?

  25. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is almost zero.

  26. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Not almost. It equals zero. Because the area of a point = 0.

  27. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    haha ..

  28. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The same situation is here. The pair of (x,y) describes the situation. So there will be a square.|dw:1350223775014:dw| Any point of this circle can show when this persons arrived. Like the shot in any point of the circle.

  29. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh. Any point of the SQUARE can show when this persons arrived.

  30. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What can you say now?

  31. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Did you get it?

  32. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no ... just on it. let me try to understand it ...

  33. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    For example, you decided to meet you friend near restaurant between 14:00 and 14:30. You can arrive at any time between 14 and 14:30 the same situation for him. Let sign the time when you arrive with x and his time - y. For example you arrives at 14:09 and he at 14:29. This will be the point |dw:1350224152495:dw|

  34. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how to represent waiting time then??

  35. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    What about \(|x-y|<\tau\) ?

  36. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1350224492870:dw| can it be independent of x?

  37. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    In my example if \(\tau=21 \) min you will meet your friend. But if \(\tau < 20\) you will not.

  38. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how to represent this probabilistically?

  39. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I thought, if first person comes after time x and waits time t then the probability is \[ {t \over T-x}\] but this is not independent of x.

  40. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    So, what about |x−y|<τ ?

  41. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    τ is not T right??

  42. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Sure!

  43. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is it (τ /T) ?

  44. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    No. Find all points (x,y) that satisfy the statement of the problem.

  45. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    pi t^2/T^2 ??

  46. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ \piτ^2 \over T^2 \]

  47. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Where did you get pi?

  48. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1350225916366:dw| sorry .. kinda thought of complex number |z| < r

  49. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The condition that they will meet is \(|x=y|<\tau\). Got it?

  50. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |x−y|<τ

  51. experimentX
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah, is that correct?

  52. klimenkov
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1350227268062:dw|

  53. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.