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#10242 11/02/05 12:32 AM
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By the way, are you using a book?

.
#10243 11/02/05 01:32 AM
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Quote:
Originally posted by Philege:
By the way, are you using a book?
No, this is a problem our prof has brought up in the first lecture on probability.

He has shown us a solution for 2 ladies, and suggested we found solution for 3 ladies problem.

If you want, I can give you a solution for 2 ladies, and you would try to solve the 3 ladies case.

e smile s

#10244 11/02/05 09:03 PM
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Okay, fire away.

#10245 11/02/05 10:33 PM
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Quote:
Originally posted by Philege: Okay, fire away.
.....I.........................5/6
.....+-----------------------+
.....|........................./:::::|
.....|....................../:::::::|.5/6
.....|.................../:::::::/...|
.....|................/:::::::/......|
.....|............/:::::::/..........|
.....|.......,/:::::::/..............|
.....|..../:::::::/..................|
1/6.|./:::::::/.......................|
.....|::::::/........................|
.....|::::/...........................|
.....+----------------------+II
.........1/6

#10246 11/02/05 10:53 PM
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Quote:
fire away.
So, the square represents all the possible combinations of arrival time for the ladies.

Only the figure in the middle marked with ":" symbols has one of the ladies waiting.

Its area is 1x1 -(5/6)x(5/6), which is 11/36.

es

#10247 11/05/05 12:20 AM
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This does not quite compute, these cannot be ALL the possible combinations, surely even you can see that! How do you come up with the dimensions of the square and the possible combinations which are so obviously wrong. Can anyone else see this? Also what about if the other woman arrive 10 minutes and one second later in each of the possible ten minutes this would represent at least 5.4 possibilies in addition to any other combinations. So I don't quite agree with your answer. Please explain in more depth. Either that or I must be really thick!

#10248 11/05/05 12:36 AM
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Quote:
Originally posted by Philege:
Can anyone else see this?
Lets say the first lady came at 1:11 and the second one at 1:25

It would be represented by the point ( 11/60 , 25/60 ) inside the 1x1 square.

Let us ask the public, if anyone agree with me?

e smile s

#10249 11/05/05 11:26 PM
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Well? We're waiting?

#10250 11/05/05 11:27 PM
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Where are ALL the fat IQs out there?

#10251 11/05/05 11:28 PM
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Doctor Sarah, Help!?

#10252 11/07/05 06:07 PM
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Quote:
Originally posted by extrasense:
Quote:
Originally posted by Philege:
Don't make me laugh, ofcourse I can
Have you ever studied probability theory, or any higher Math at all?
Here is a little probability problem:
There is small bench that can accomodate two, overseeing a pond - in the park . Three ladies decided to come and sit there for 10 minutes, between 1pm and 2pm. They come there independently. What are the chances, that one of them will have to wait because bench is occupied by other two?

e smile s
2/27

#10253 11/08/05 12:02 AM
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Quote:
Originally posted by Count Iblis II:
2/27
This is correct. Congratulations!

How did you solve it?

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#10254 11/08/05 12:35 AM
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Here is how I did it (I tried to post this before but I got an error message due to the symbol for ''smaller'' or ''larger'' being recognised as an illegal HTML command.):

Write the probability as an integral over the arrival times denoted as t1, t2 and t3 for the three ladies, respectively. Imposing an ordering t1 smaller than t2 smaller than t3 would reduce the probability by a factor of 6 by symmetry so you can solve theproblem using this ordering and then multiply by 6. Then you split the problem for t1 below 5/6 and t1 between 5/6 and 1. In the first case t2 goes from t1 to t1 + 1/6 and t3 goes from t2 to t1 + 1/6 and in the second case t2 goes from t1 to 1 and t3 goes from t2 to 1.

To simplify the calculations you partially undo the imposed ordering that led to the factor 6. If you allow t2 to be come larger than t3 (i.e. you add contributions in the two cases for t2 and t3 interchanged), then you must replace the factor 6 by a factor 3 and the two cases become:

1) t2 goes from t1 to t1 + 1/6 and t3 goes from t1 to t1 + 1/6

2) t2 goes from t1 to 1 and t3 goes from t1 to 1.

The first case yields a probability of 5/6*1/6*1/6*3

The second case yields 1/3 (1/6)^3 * 3

Which gives the result of 2/27

#10255 11/08/05 02:40 AM
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Quote:
Originally posted by Count Iblis II:
Write the probability as an integral over the arrival times denoted as t1, t2 and t3 ...
This is a solution that requires, oh horror! eek , integrals!
Nevertheless, it was not that bad, was it?

Best,

e smile s

#10256 11/08/05 10:43 PM
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Ehem! But extrasense you said you would show me the solution for this, what happened?

#10257 11/08/05 10:45 PM
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Thanks count Iblis II.

#10258 11/09/05 12:02 AM
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Quote:
Originally posted by Philege:
Thanks count Iblis II.
What is good about count's solution, is that it can be easily extended to 4, 5, .. ladies. And the gents too

e :p s

#10259 11/09/05 09:07 AM
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Hi, can anybody say me, if programs and sites for IQ training are really useful?
I've just downloaded software for iq training from http://www.soft32.com/download_125585.html ,
however i don't know, if it is really possible to increase IQ up to 40% as they say.
I mean the following- if my current IQ is 100 (for example), after 3 months of training it will be 140. Ok, I'll start training again, and my IQ will increase up to 196. I'll train more and more- and finally my IQ score will be more than 1000 smile
How large IQ score scale actually is, does anybody know?

#10260 11/09/05 11:29 AM
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I think it's limitless

#10261 11/09/05 10:50 PM
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I think it's limitless


I don't think so, once you go over the brink, your IQ starts to reverse, and you know what that means.

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