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#1 2015-12-05 20:50:56

mr.wong
Member
Registered: 2015-12-01
Posts: 252

geometric probability--segments

Inside a segment E there are 3 segments A, B, and C which move freely along E, and all with lengths being 1/2 of E. If a point is chosen randomly on E, find the probability that the point lies within A, B, and C at the same time.

There are 2 possible answers to this problem:
(1)1/4
(2) about 1/5
I am not sure which one is correct,can the correct answer be verified by using computer?

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#2 2015-12-05 20:54:17

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

It can be verified by computer, I assume the 3 line segments can overlap and I assume that the 3 segments are all entirely inside E. I am getting 1 / 4.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2015-12-06 15:06:48

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Thank you very much bobbym !  I am looking to find why the 2nd answer is incorrect.

If there are only 2 smaller segments ( say A and B only) inside E,will you get  P= 1/3 ?

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#4 2015-12-06 16:12:16

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Yes, I believe you will.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2015-12-07 15:05:39

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Thanks again bobbym!

Now let  n  denotes the no. of smaller segments inside E( each with length being 1/2 of E). For n=1, surely P  = 1/2 ; for n=2, we get P= 1/3 ; for n=3, you get P=1/4 ; for n=4, will you guess P= 1/5  and so on ?

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#6 2015-12-07 15:37:38

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Earl wrote:

Math is a beautiful instrument played by ugly people.

I have verified P=1/5 and P = 1/6 experimentally.

and so on

I have enough empirical evidence to conjecture that for n, P(n) =1 / (n+1).


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#7 2015-12-08 20:43:57

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Much thanks to bobbym !

Previously I had never thought there may be such a relation between  n  and  P. I will try to
prove that formula by mathematical induction .

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#8 2015-12-08 20:53:59

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

When you prove it please post it here.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2015-12-09 17:06:04

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Hi ! bobbym

I think I am unable to prove the formula. I cannot even deduce  P(2) to P(3),that's why I got a wrong answer for P(3).
However,somebody can solve P(3) by multiple integration and get a correct answer. Thus the formula may be proved by mathematical induction with assistance of multiple integration.

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#10 2015-12-09 18:52:17

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

I am unable to come up with a proof for a general form.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#11 2015-12-21 16:38:25

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Assume that the formula  P(n)= 1/n+1  is true, then it can further be generalized .

Let  P(n,r) denotes the probability that the point lies within exactly  r  smaller segments out of the  n ,(where r varies  from  0  to  n ) then  P(n,r) ≡ 1/n+1 ,i.e. no matter what is the value of  r , P(n,r) always = 1/n+1 .
Since there are  nCr  items  for each  r , if their total  P  sum up to  1/n+1 ,then  P  of each item will be 1/[(n+1)(nCr )]. ( The  total no. of various items will be  nC0 + nC1 + nC2 + ........+nCn  =  2 ^ n .)
E.g. ,let  n = 3,
for  r = 0, there is only 1 such item,( neither  A nor B nor C ) and P(3,0) = 1/4;
for  r = 1, P(3,1) also = 1/4 ,since there are  3C1 = 3 such items ,( either  A  or  B  or C ),thus each item will get
                 P = 1/4 * 1/3 = 1/12 ;
for  r  = 2, P(3,2) also = 1/4 , since there are 3C2 = 3 such items ,( either  A with B ,or A with C , or B with C ),
                 thus each item will get  P = 1/12  also ;
for  r = 3, there is only 1 such item ( A with B with C ) and  P (3,3) = 1/4 .

Last edited by mr.wong (2015-12-21 16:41:50)

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#12 2015-12-21 17:21:05

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

I am sorry that I did not post this earlier, it is from a discussion of this problem that took place in another thread. You might find it interesting...

http://forumgeom.fau.edu/POLYA/ProblemC … LYA034.pdf


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#13 2015-12-24 16:58:04

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Hi bobbym ,


The  2  problems  look  like quite  similar but  in  fact  they are  quite  different . In  that  problem  the  probability of  the  last  hit  depends on  the  result  of  the  previous  throws , but  in  my  problem  the  probability  of  the point  lies within  each  segment  is  fixed  .( Each  segment  has  same  length .) Moreover , in  that  problem n  has  to  start  from  3  but  in  mine  n  can  be  started  from  1 .
However , I  am  amazed  that  the  results  of  them  will  be  exactly  the  same  if  one  of  them  is  shifted
by  2  positions .

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#14 2015-12-24 17:39:48

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

I agree, I do not now see any similarity between the 2 problems.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#15 2015-12-27 15:07:53

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

In  fact  I  am  more  interested  to find  a  formula  for  problems  involving  3  smaller  segments  with  their  individual  lengths  other  than  1/2  with  respect  to  e ( length  of  E ) .
Let  P ( a, b, c )  denotes   the  probability ( estimated ) of  a  point  chosen  in  E  lies  within  A ,B , and  C with their lengths  being  a, b, and c  respectively ,( where  a , b, and  c may  be  values  other  than  1/2 , say  from  1/10  to  9/10 ,for  various  a,b, and c , i.e. their  values   may  be  different  .)
If  I  can  obtain  enough such data  of  P ( a, b, c)  ,perhaps  I  can  guess  a  general  formula  for  it .

Last edited by mr.wong (2015-12-27 15:09:29)

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#16 2015-12-27 15:35:42

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

You want it always to be 3 smaller segments and you want the lengths to vary?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#17 2015-12-28 21:23:03

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Hi  bobbym ,

Yes , that's  what  I  want . Can  you  give  a  help ?

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#18 2015-12-29 00:32:39

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

I can generate some more data and attempt to use the methods of Experimental Math to derive a formula. But even with lots of data this is sometimes impossible. I will try though.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#19 2016-01-10 20:47:38

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

The first bit of real results in a long time of hard work.

If we denote the length of the three segments that will slide along E, S1,S2 and S3.

1) Then:

That covers some of the cases, I have more but need time to check them...


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#20 2016-01-11 17:03:46

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Hi  bobbym ,

Thank  you  very  much  for  your  hard  working , but  don't  spend  too  much  spirit  by  your  way , it  does  not  worth !
Please  share  some  original  data  with  me ! I  can  do  it  much  more  concisely since  I  already  have  a  formula  for  2  smaller  segments .
I need  time  to  analyse  your  formula , but  what  can  I  do  will  be  limited  unless  I  have  enough  original  data  in  hand .

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#21 2016-01-12 00:28:06

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

I did not need much data to generate the formulas so I will post what I have till I generate more. Would you please post your formula for 2 segments?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#22 2016-01-14 21:48:00

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Hi  bobbym ,

I  recognize  that  I  can't  derive  a  formula  for  3  segments  from  your  data .

The  following  is  the  problem  involving  2  smaller  segments  and  it's  formula  .

Problem :

Inside  a  segment  E ( with  length  being  1  unit ) there  are  2  smaller  segments  A  and
B  ( with respective  lengths   a  and  b  units  , where  0≦ a ≦ 1   and  0 ≦ b ≦ 1)  which  slide  freely  along  E .
(1) Find  the  expectation  of  the  length  of  common  portion  of  A  and  B .
(2) If  a  point  is  chosen  randomly  on  E , find  the  probability  that  the  point  lies
      within  the  common  portion  of  A  and  B .

Formula  for  (1) : 
min ( a,b ) -   [ min ( 1-a, 1-b )^3 - max ( 0, 1-a-b )^3 ] / 3(1-a)(1-b)  unit
where  min  denotes  the  smallest  value  while  max  denotes  the  greatest  value .

By  dividing  the  expression  by  1  unit , we  get  the  answer  of (2)  readily .

( Welcome  to  check  the  validity  of  the  formula  by  computer .)

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#23 2016-01-15 00:21:10

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

It is somewhat simpler to avoid the max and min and use

Have not tested this much but it seems ok.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#24 2016-01-16 15:54:35

mr.wong
Member
Registered: 2015-12-01
Posts: 252

Re: geometric probability--segments

Hi  bobbym ,

I  think  your  formula  may  be  correct    for  certain  limited  cases , but  not  always . Also  the  formula  should  be  symmetric  for  a  and  b .
Moreover , if  both  a  and  b  are  small  enough , the formula  may  yield  a  negative  value !

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#25 2016-01-16 17:27:57

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: geometric probability--segments

Hi;

It does have some conditions on it, can I see your examples where it fails?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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