geometric probability--segments

mr.wong · 2015-12-05 20:50:56

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?

bobbym · 2015-12-05 20:54:17

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.

mr.wong · 2015-12-06 15:06:48

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 ?

bobbym · 2015-12-06 16:12:16

Yes, I believe you will.

mr.wong · 2015-12-07 15:05:39

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 ?

bobbym · 2015-12-07 15:37:38

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).

mr.wong · 2015-12-08 20:43:57

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 .

bobbym · 2015-12-08 20:53:59

When you prove it please post it here.

mr.wong · 2015-12-09 17:06:04

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.

bobbym · 2015-12-09 18:52:17

Hi;

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

mr.wong · 2015-12-21 16:38:25

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)

bobbym · 2015-12-21 17:21:05

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

mr.wong · 2015-12-24 16:58:04

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 .

bobbym · 2015-12-24 17:39:48

Hi;

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

mr.wong · 2015-12-27 15:07:53

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)

bobbym · 2015-12-27 15:35:42

Hi;

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

mr.wong · 2015-12-28 21:23:03

Hi bobbym ,

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

bobbym · 2015-12-29 00:32:39

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.

bobbym · 2016-01-10 20:47:38

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...

mr.wong · 2016-01-11 17:03:46

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 .

bobbym · 2016-01-12 00:28:06

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?

mr.wong · 2016-01-14 21:48:00

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 .)

bobbym · 2016-01-15 00:21:10

Hi;

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

Have not tested this much but it seems ok.

mr.wong · 2016-01-16 15:54:35

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 !

bobbym · 2016-01-16 17:27:57

Hi;

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

Math Is Fun Forum

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

geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

Re: geometric probability--segments

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

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

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

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