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#51 2016-04-06 21:59:18

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

Re: geometric probability ---- squares

Thanks  bobbym !

Thus  the  polyhedron  has  an  apex  at  the  point
with  co-ordinate  (1/6 , 1/6 , 1/12) .
Waiting  for  a  solution  with  integration , let  us  firstly 
try  to  find  an  approximate  answer  of  the  problem  with  geometry .
Since  the  curve  over  QR  = 1/2 x - xx ,
we  can  find  the  area  under  the  curve
from  0  to  1/2  by  using  simple  integration
and  obtain  1/48  sq.un., thus  the  average
height ( representing  the  average  common  area  of  A  and  X )
= (1/48)  / ( 1/2) = 1/24 .
Similarly  the  average  height  over  PQ  also  = 1/24 .
We  have   omitted  the  portion  of  the  polyhedron  higher  than  1/16 
to  the  apex  as  its  value  is  so  small  and  will  not  affect  much .
Divided  1/24 sq.un  by the  area  of Δ ABC , i.e.  1/8  sq.un., we  get   
the  probability  being  about  1/3 .

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#52 2016-04-08 03:10:13

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

Re: geometric probability ---- squares

Hi;

Which question is that answering?


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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#53 2016-04-08 15:09:09

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

Re: geometric probability ---- squares

Hi  bobbym ,

That  answer  is  for  Problem  (III B)  at  #  46 .

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#54 2016-04-09 06:55:03

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

Re: geometric probability ---- squares

Hi;

I will check to see if I can get the same answer but that will be later because I am being called away for now.


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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#55 2016-05-17 20:29:20

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

Re: geometric probability ---- squares

Hi  bobbym ,

Finally  I  obtained  the  volume  of  the  polyhedron  to  be  1/128 .
Thus  its  average  height  ( representing  the  average  value  of  common
area ) =  (1/128 ) / (1/8) = 1/16 . Therefore  the  probability  required
= ( 1/16) / (1/8) = 1/2 . ( for  Problem  (III B)  at  #  46 )
Does  this  value  coincide  with  your  simulation ?

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#56 2016-05-17 20:58:20

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

Re: geometric probability ---- squares

Hi;

It does not coincide with what I had in my notes and I will need to rethink the whole problem.

My answer was 1 / 12 which came from a simulation and your maximized function in post #50.

Do you still want to use the diagram in post #46?


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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#57 2016-05-18 00:12:36

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

I have got an idea which I am just initiating. For simplicity we can take the vertices as O(0,0) A(a,0) and B(0,b). let P be point (x,y) at which we want probability.Imagine the moving triangle with its vertices in their turn at P. say CDP where P corresponds to A of

we get C(x-a/2,y+b/2) D(x-a/2,y) Similarly for other positions E(x,y-b/2) F(x+a/2,y-b/2) G(x+a/2,y) H(x,y+b/2). the area CDEFGHC is the one where P is inside the moving triangle but only one condition, the hexagon must not step out of triangle ABC. this area divided by area of triangle ABC should give the probability distribution which when integrated over the triangle joining midpoints of  triangle ABC gives the total probability. How to solve it  I have not analyzed so far.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#58 2016-05-18 08:15:28

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

Re: geometric probability ---- squares

Hi mr.wong;

I am getting an answer of 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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#59 2016-05-18 17:06:54

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

Re: geometric probability ---- squares

Hi  bobbym ,

I  think  the  diagram  in  post  #46  is  still  useful .
I   am  not  sure  that  my  answer  of  1/2  is  correct ,
I  will  keep  on  investigating .

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#60 2016-05-18 17:08:09

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

Re: geometric probability ---- squares

Hi  thickhead ,

Thanks  for  your  reply !  But  I  don't  quite  understand
your  statement . It  would  be  much  helpful  if  you 
have  a  diagram   or  sketch  with  the  post .
For  simplicity  just  let  a = b = 1/2  in  this  stage ,
and  generalize  the  result  if  everything  comes  ok .
Integration  is  always  useful  in  calculating  areas  and 
volumes , but  I  am  weak  in  calculus . Wish  to  receive  your  help  in  problem  dealing  with  integration  and  other  related  fields .

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#61 2016-05-18 17:11:55

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

Re: geometric probability ---- squares

If I am interpreting the problem correctly the answer is 1 / 4 as stated above. This fits a simulation as well as an analytical solution.


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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#62 2016-05-19 01:22:31

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

I was referring to your problem III I am getting the probability as 7/16 with the method I laid out before this. Did you come across this value sometimes in your analysis? I do not have the technique to give the diagram but I can just describe. I am taking the vertices of the outer triangle as O(0,0) A(2,0) and B(0,2) the midpoints are L(1,0)M(1,1) and N(0,1).just to have a glimpse of the hexagon I described before  take a random pt P(x,y) at (0.8,0.8)
Then plot these pts C(-0.2,1.8) D(-0.2,0.8) E(0.8,-0.2) F(1.8,-0.2) G(1.8,0.8) and H(1.8,1.8) If the moving triangle is within this hexagon th point (x,y) is inside the triangle but part of the hexagon going out of triangle OAB is invalid. Only the common area of the hexagon and triangle OAB is valid. From the diagram we find 3 small triangular portions near the vertices are the left out portions. The fraction of sum of these areas over the area of triangle OAB when subtracted from 1 gives the probability that (x,y) lies inside the moving triangle. This expression has to be integrated over the area LMN gives the average probability.Don't consider (0.8,0.8) in any calculation. It is used just to draw a typical diagram.

Last edited by thickhead (2016-05-19 01:23:52)


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#63 2016-05-19 04:46:39

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

mr.wong wrote:

Inside  a  square  E  there  are  2  smaller  squares  A  and  B  ,both with  length  of  sides  being 1/2  of  that  of  E.  Both  squares
can  move  freely  inside  E,  but  must  keep  "parallel " with  E.  If  a point  is  chosen  randomly  on  E , find  the  probability  that  the  point lies  inside  A  and  B  at  the  same  time.

I am getting the  average probability of a point  to be inside one of the squares as 9/64 and inside both as 81/4096 using the method as given previously.


Now this is found to be incorrect.

Last edited by thickhead (2016-06-18 22:16:39)


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#64 2016-05-20 15:08:58

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

Re: geometric probability ---- squares

Hi  bobbym ,

You  got  an  answer  of  1/8  for  Problem ( III) in  #  38 ,  this  is 
equivalent  to   an  answer  of  1/2  for  Problem  (III)( B ) in
# 46 .  I  wonder  why   you  give  up  this  result ?
Though  triangle  X  occupies  1/4  of  the  area  of  triangle  E ,
but  its  position  is  at  the  centre  of  E , thus  the  probability 
that  it  overlaps  with  triangle  A  should  be  greater .

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#65 2016-05-20 16:45:43

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

mr.wong wrote:

Inside  a  square  E  there  are  2  smaller  squares  A  and  B  ,both with  length  of  sides  being 1/2  of  that  of  E.  Both  squares
can  move  freely  inside  E,  but  must  keep  "parallel " with  E.  If  a point  is  chosen  randomly  on  E , find  the  probability  that  the  point lies  inside  A  and  B  at  the  same  time.

Hi wong,
I have been attracted by your problems recently.I propose to use a diagram which I may christen as "favourable mobility diagram" if it is not already existing. that is why I start with the simplest problem to check it and go to complicated problems later. irequest you and Bobbym to critically examine the same and correct me if and when necessary.
In this problem, consider a point P(x) in the left half of E. Way of approach is different for the right half but the results will be same. Consider the position of A when its right edge is flush with P. Consider another position of A with its left edge flush with P.Combine the two.This gives a diagram for the position of A when P is inside A.This I want to call "Favorable mobility diagram". Part of this is outside E; that part is invalid.Therefore the valid area (which is inside E also) is x+1/2 but out of this area 1/2*1 is self occupied area of A which is not available for mobility. Therefore only area x (i.e. x*1)is favorable area whereas total mobility is the area of E-area of A (again self occupied area) which comes out to be 1/2. therefore probability of P being inside A =x/(1/2)=2x.
Now let us compare it with analytcal approach. Consider position of A with its left edge flush with left wall of E. Now P is inside A. Now start moving A towards right. when its left edge clears p P starts falling outside A. So favorable distance is x. the triangle A however can move a maximum distance of 1/2.So the probability is x/(1/2)=2x. I hope this convinces the use of "favorable mobilty diagram"
Now coming to the solution of the problem,previously i had made the mistake of integrating this expression to get average probability for A and multiplied the result with itself for the combined probability of P being inside both A and B.this is incorrect. what one should do is at P itself multiply the 2 probabilities and integrate over the region.
Hence

Last edited by thickhead (2016-05-20 16:46:28)


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#66 2016-05-20 17:23:08

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

Re: geometric probability ---- squares

mr.wong. wrote:

You  got  an  answer  of  1/8  for  Problem ( III) in  #  38 ,  this  is
equivalent  to   an  answer  of  1/2  for  Problem  (III)( B ) in
# 46 .

Problem  (III)( B ): I meant the expected value of the overlapping area of the two triangles is 1 / 4. I am getting a probability of 1 / 2.

mr.wong: I have been solving all the problems based on the assumption the two smaller triangles and the two squares are fully inside the bigger triangle and square. No part of these triangles or squares is outside the larger triangle or square. Do you agree?

Hi thickhead;

I can not work on these anymore until mr.wong answers the above question.


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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#67 2016-05-20 18:38:53

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

mr.wong wrote:

Thanks  bobbym , you  are  right .

By  formula , we  have  the  probability  of  the  point  lies
within  the  axis  of 1  side ( say  horizontal  side ) of  the
common  portion  of  A  and  B  to  be  1/3 . Similarly ,
for  the  vertical  side , P  also  =  1/3 . Thus  combinely
P  of  the  point  lies  within  both  A  and  B  will  be  1/9 .
Will  the  answer  be  the  same  for  other  polygons , say
rhombus  under  similar  conditions ?

Could you elaborate on the formula?


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#68 2016-05-20 19:32:41

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

mr.wong wrote:

Inside  a  square  E  there  are  2  smaller  squares  A  and  B  ,both with  length  of  sides  being 1/2  of  that  of  E.  Both  squares
can  move  freely  inside  E,  but  must  keep  "parallel " with  E.  If  a point  is  chosen  randomly  on  E , find  the  probability  that  the  point lies  inside  A  and  B  at  the  same  time.

For the purpose of clarity I take E to be of size 2*2 and A and B to be 1*1.
Consider a point P(x,y) 0<x<1,0,y<1; I bring the 4 vertices of A to point P one at a time to form my "favorable gross mobility diagram" I have added gross because self occupied area of a is to be subtracted from it. Now part of this diagram inside E is a rectangle of sides x+1 and y+1. Its area is (x+1)*(y+1). To get net favorable area we have to subtract area of A viz. 1*1. So the favorable net mobility area =(x+1)(y+1)-1=xy+x+y. To get the probability of P lying inside A we have to divide it total net mobility area=area of E-area of A=2*2-1*1=3.
So

  We can check the value of p at selected points. At (0,0) ,p=0 and at (1,1) p=1;This is correct because at (0,0) just one corner of A coincides with it just for a fraction of moment and (1,1) will be inside or on A for all the time. the same value applies for B. Integrating we get,

Last edited by thickhead (2016-05-20 20:38:01)


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#69 2016-05-20 20:06:57

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

Re: geometric probability ---- squares

Hi  bobbym  and  thickhead ,

From  the  beginning  , I  have  assumed  that  " the two smaller triangles and the two squares are fully inside the bigger triangle and square. No part of these triangles or squares is outside the larger triangle or square."
I  hope  this  will  help  to  clarify  the  problems .

Hi  thickhead ,

I  need  time  to  read  through  your  several  posts .
  I  shall  reply  later  if  I  understand  your  work .

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#70 2016-05-20 20:32:58

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

Re: geometric probability ---- squares

That is how I saw it too but as there is some disagreement in the 2 square in the bigger square problem maybe we should see if we can all agree on the answer...


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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#71 2016-05-20 21:41:49

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

I have understood wong's condition that the 2 smaller squares are always within the bigger triangle. My "favourable mobility diagram" is virtual and any of its part lying outside the bigger square is to be discarded.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#72 2016-05-23 15:08:29

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

Re: geometric probability ---- squares

Hi  thickhead ,

For  the  problem  involving  the  squares , it  seems  you  have 
considered  the  area  which  has  been  covered  by  square  A 
during  moving , but  not  the  area  being  covered  by  A  after 
moving  . This  is  not  the  aim  of  the  original  problem .
Even  so  the  " gross  mobility  diagram  "  should  looks  like 
a  hexagon  instead  of  a  rectangle . ( if  moving  in  a  straight 
line )  Thus  the  net  area  should  be  less  than   xy + x + y .
I  don't  know  whether  my   comprehension  of   your  statement 
is  right  or  not . Please  correct  me  otherwise .
Nevertheless , I  still  persist  that  with  the  aim  of  the  original 
problem , the  answer  should  be  1/9 .

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#73 2016-05-23 17:07:01

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

In the case of moving square the gross mobility diagram is also a square of double length and double height. It is not a hexagon. The idea is Take the moving square and slide all  its borders over point P(x) so that P is just within the square. In the case of triangle it turned out to be a hexagon. The area of overlap of "gross mobility diagram" with the outer square( which I have taken as 2*2) comes out to be (1+x)*(1+y) From this I have subtracted  1*1  the area of the mobile square. This is the net area of mobility during which P remains inside/just on border of moving square A. the overall mobility of square a is 2*2-1*1=3. hence the probability p is calculated as (xy+x+y)/3. with 2 squares A and B probability would be p^2 and this is to be integrated over the area of south_west 1x1 square.Over the other squares the expression for p would be different and limits of integration would be different but will yield the same result.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#74 2016-06-08 04:45:51

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: geometric probability ---- squares

Last edited by thickhead (2016-06-08 16:20:56)


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#75 2016-06-08 15:45:50

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

Re: geometric probability ---- squares

Thanks  thickhead ,

I  agree  with  your  results . For  n moving  squares  the 
probability  seems  to  be  1/ (n+1)^2 . ( but  not  yet  proved )

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