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Hi thickhead ,
Can you confirm your answer ( 1/10 ) of E involving 2 moving triangles
by getting the result directly with multiple integration other than indirectly deduced from results of the 4 smaller triangles of E ?
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The local probability function ( depending on the position (x,y) of the point ) differs for the 4 smaller triangles. I have to evaluate them separately and combine.
{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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Hi thickhead ,
Will you expect apart from isosceles right angle triangles ,
the results will remain the same for any other triangles , say
equilateral triangles ?
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There is a demonstration for any type of triangle as long as all your conditions apply.
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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Hi bobbym ,
If the theory of that demonstration is too hard for me ,
then I shall skip it .
In the following we shall try to deduce some related results from
what we have got .
With reference to the original post in this thread :
Let P ( A | E ) denotes P of the point chosen in E which also lies in A .
and P ( ∼A | E ) denotes P of the point chosen in E which lies outside A
being 1- 1/4 = 3/4 .
(1) Let P ( A ∩ B | E ) denotes P of the point lies inside both A and B [ the intersection of A and B ] , with value being 1/10 ( according to thickhead 's result ) , thus P ( ∼A ∩ B | E ) = P ( B | E ) - P ( A ∩ B | E ) = 1/4 - 1/10 = 3/20 . being P of the point lies outside A but inside B .
( Notice that it should not be calculated as P ( ∼A | E ) * P ( B | E ) =
3/4 * 1/4 = 3/16 .)
(2) Since P ( A ∩ B ∩ C | E ) = 1/21 ( thickhead's result ) , thus
P ( ∼A ∩ B ∩ C | E ) = P ( B ∩ C | E ) - P ( A ∩ B ∩ C | E )
= 1/10 - 1/21 = 11/210 being P of the point lies outside A but inside
both B and C .
(3) P (∼ A ∩∼ B ∩ C | E ) = P ( C | E ) - P ( A ∩∼ B ∩ C | E ) -
P ( A ∩ B ∩ C | E ) - P ( ~ A ∩ B ∩ C | E ) = 1/4 - 11/210 - 1/21 - 11/210
= 41/ 420 ( may be illustrated more clearly with a Venn Diagram )
Related problem (I) :
Find the probability of a point chosen in E randomly which lies within
one and only one moving triangle . ( answer : 41/ 140 )
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Will you expect apart from isosceles right angle triangles ,
the results will remain the same for any other triangles , say
equilateral triangles ?
It at least shows that this is true provided his work is correct.
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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For 2 moving triangles in E
Last edited by thickhead (2016-06-17 03:37:46)
{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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Thanks thickhead ,
Elegant and laborious work !
Since the common portion of 2 similar and parallel triangles
is still a similar and parallel triangle ( This is not true for squares ) ,
originally I had expected that the answer should be the square of certain
average ratio of the corresponding sides of overlapped triangle with E .
In fact it is much more complicated .
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Once the probability functions have been derived you can even get for 10 moving triangles just by using these integrations but with power of 10 but it is not possible for "n" moving triangles as the integrals are not that easy for general terms. In case of squares it was possible for n squares. here also the troublesome is the integration in triangle X, the others can be managed.
{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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Related problem (II) :
If a point is chosen randomly in square LQMN , find the probability
that the point lies inside at least 2 moving triangles .
( answer : 32/105 )
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Hi bobbym ,
If the theory of that demonstration is too hard for me ,
then I shall skip it .
In the following we shall try to deduce some related results from
what we have got .
With reference to the original post in this thread :
Let P ( A | E ) denotes P of the point chosen in E which also lies in A .
and P ( ∼A | E ) denotes P of the point chosen in E which lies outside A
being 1- 1/4 = 3/4 .
(1) Let P ( A ∩ B | E ) denotes P of the point lies inside both A and B [ the intersection of A and B ] , with value being 1/10 ( according to thickhead 's result ) , thus P ( ∼A ∩ B | E ) = P ( B | E ) - P ( A ∩ B | E ) = 1/4 - 1/10 = 3/20 . being P of the point lies outside A but inside B .
( Notice that it should not be calculated as P ( ∼A | E ) * P ( B | E ) =
3/4 * 1/4 = 3/16 .)
(2) Since P ( A ∩ B ∩ C | E ) = 1/21 ( thickhead's result ) , thus
P ( ∼A ∩ B ∩ C | E ) = P ( B ∩ C | E ) - P ( A ∩ B ∩ C | E )
= 1/10 - 1/21 = 11/210 being P of the point lies outside A but inside
both B and C .
(3) P (∼ A ∩∼ B ∩ C | E ) = P ( C | E ) - P ( A ∩∼ B ∩ C | E ) -
P ( A ∩ B ∩ C | E ) - P ( ~ A ∩ B ∩ C | E ) = 1/4 - 11/210 - 1/21 - 11/210
= 41/ 420 ( may be illustrated more clearly with a Venn Diagram )
Related problem (I) :
Find the probability of a point chosen in E randomly which lies within
one and only one moving triangle . ( answer : 41/ 140 )
Is this your answer or required correct answer?
Alternate way:
Probability of point lying in A but not B &C=
Last edited by thickhead (2016-06-20 16:43:44)
{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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Hi thickhead ,
You got the same answer of 41/420 with me .
(where the term 2/20 should be 2/10 ) But sometimes
it may cause confusion since P stands for 1/4 ( thus
( 1-P ) should be 3/4 ) while in the expression
P-2P^2 + P^3 , P^2 stands for P for 2 moving
triangles ( instead of 1/4 * 1/4 =1/16 ) and P^3 stands for P for 3 moving triangles .( instead of 1/4* 1/4 * 1/4 =1/64 ) .
But I wonder why it seems you can always get a correct answer by your way ! ( including P for triangle A or triangle X )
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Last edited by thickhead (2016-06-20 16:48:26)
{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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Related problem (II) :
If a point is chosen randomly in square LQMN , find the probability
that the point lies inside at least 2 moving triangles .
( answer : 32/105 )
I presume there are 3 moving triangles. i am getting answer as 213/2688. is your answer reliable?
{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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{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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Hi thickhead ,
For square LQMN I mean the square taken from
triangle P (0,2) ,Q (0,0) R (2,0) , thus its coordinates
will be L ( 0,1) , Q (0,0) ,M ( 1,0) and N (1,1) . So a diagram is always helpful !
Should L be (0,2) in your problem ?
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I thought triangle E is doubled and you renamed it LQMN. Yes L (0,2) I had taken. Not only that I had to derive 8 probability functions separately. may be useful in future.
Last edited by thickhead (2016-06-21 16:53:42)
{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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{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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Related problem :
Let E denotes a triangle PQR with PQ = QR = 6 units and
angle Q = 90 degree . Inside E there are 2 similar triangles
A and B both being parallel with E and can move freely and
uniformly inside E but must keep parallel with E in moving .
The lengths of the corresponding sides of A and B are
3 units and 2 units respectively .
If a point is chosen randomly on E , find the probability that
the point also lies inside A and B at the same time .
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Hi thickhead ,
Will the property of conjugates also applies to triangles ?
For example , if the length of the corresponding side of
B is changed to 4 units , will the corresponding probability be double ?
or something like that ?
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In triangles so far we have dealt with moving triangles whose sides are 1/2 of stationary triangle. Before attempting your problem of #44 I want to try the problem with triangles A ,B and Cwith sides=2 only.
{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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An interesting point with triangle E with vertices (0,0) (1,0)and (1,0) and moving triangle A of side a is that
(1) For a<1/3 there is a central triangle of sides 1-3a (with running track of width
Last edited by thickhead (2016-08-02 16:21:37)
{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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Hi thickhead ,
For (1) should it be a < 1/3 ?
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Hi mr.wong,
You are right. I corrected it.
{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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