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Probability of chossing two points from a segment of length L so that one is 2L/3 greater than another.

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2

$begingroup$

Two points are selected randomly on a line of length $L$ so as to be on the opposite sides of the midpoint of the line. In other words, two points X and Y are independent random variables such that X is uniformly distributed over $(0,L/2)$ and Y is so over $(L/2,0)$. Find the probability that the distance between these two points is greater than $2L/3$.

Here's what I have done.

Can I get some help to finish this.

probability probability-theory conditional-probability

share|cite|improve this question

edited Mar 21 at 14:32

Ethan Bolker

45.6k553120

asked Mar 21 at 14:24

ChakSayantanChakSayantan

492511

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not help finishing but a suggestion on how to start. You can assume $L=2$. Then find the area of the part of the unit square that satisfies your condition. (This may in fact be what you've done, with the picture making the calculations clearer,.)
  $endgroup$
  – Ethan Bolker
  Mar 21 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah. Quite good idea. Thanks.
  $endgroup$
  – ChakSayantan
  Mar 21 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $L=2$ and define $A=1-X$, $B=Y-1$, then $Y-X=A+B$. $A$ and $B$ are then i.i.d. over $(0,1)$ and you are looking for $P(A+Bgefrac43)$
  $endgroup$
  – Daniel Mathias
  Mar 21 at 14:52
  

  
  
  
  

add a comment | 

2

$begingroup$

Two points are selected randomly on a line of length $L$ so as to be on the opposite sides of the midpoint of the line. In other words, two points X and Y are independent random variables such that X is uniformly distributed over $(0,L/2)$ and Y is so over $(L/2,0)$. Find the probability that the distance between these two points is greater than $2L/3$.

Here's what I have done.

Can I get some help to finish this.

probability probability-theory conditional-probability

share|cite|improve this question

edited Mar 21 at 14:32

Ethan Bolker

45.6k553120

asked Mar 21 at 14:24

ChakSayantanChakSayantan

492511

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not help finishing but a suggestion on how to start. You can assume $L=2$. Then find the area of the part of the unit square that satisfies your condition. (This may in fact be what you've done, with the picture making the calculations clearer,.)
  $endgroup$
  – Ethan Bolker
  Mar 21 at 14:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ah. Quite good idea. Thanks.
  $endgroup$
  – ChakSayantan
  Mar 21 at 14:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Let $L=2$ and define $A=1-X$, $B=Y-1$, then $Y-X=A+B$. $A$ and $B$ are then i.i.d. over $(0,1)$ and you are looking for $P(A+Bgefrac43)$
  $endgroup$
  – Daniel Mathias
  Mar 21 at 14:52
  

  
  
  
  

add a comment | 

$begingroup$

Two points are selected randomly on a line of length $L$ so as to be on the opposite sides of the midpoint of the line. In other words, two points X and Y are independent random variables such that X is uniformly distributed over $(0,L/2)$ and Y is so over $(L/2,0)$. Find the probability that the distance between these two points is greater than $2L/3$.

Here's what I have done.

Can I get some help to finish this.

probability probability-theory conditional-probability

share|cite|improve this question

edited Mar 21 at 14:32

Ethan Bolker

45.6k553120

asked Mar 21 at 14:24

ChakSayantanChakSayantan

492511

$endgroup$

share|cite|improve this question

edited Mar 21 at 14:32

Ethan Bolker

45.6k553120

asked Mar 21 at 14:24

ChakSayantanChakSayantan

492511

share|cite|improve this question

edited Mar 21 at 14:32

Ethan Bolker

45.6k553120

asked Mar 21 at 14:24

ChakSayantanChakSayantan

492511

edited Mar 21 at 14:32

Ethan Bolker

45.6k553120

edited Mar 21 at 14:32

Ethan Bolker

45.6k553120

asked Mar 21 at 14:24

ChakSayantanChakSayantan

492511

asked Mar 21 at 14:24

ChakSayantanChakSayantan

492511

2 Answers
2

active

oldest

votes

3

$begingroup$

Does this make sense to you? I think this aligns with what Ethan has commented.

share|cite|improve this answer

answered Mar 21 at 14:42

J. WangJ. Wang

1916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah. I got it.
  $endgroup$
  – ChakSayantan
  Mar 21 at 14:47
  

  
  
  
  

add a comment | 

0

$begingroup$

A picture, as provided by J. Wang, is very useful for intuition (and has my +1!), but it does need to be made into a rigorous proof. My suggestion is to try conditioning on the first point, $ell_1$. Once you know this, you can ask "Which options for $ell_2$ give the required condition?"---this clearly depends on $ell_1$.

• If $ell_1 > L/3$, then clearly no $ell_2$ will suffice.


• If $ell_1 in [0,L/3]$, then any $ell_2 in [ell_1+2L/3,L]$ will suffice.

Given $ell_1$, and that we land in the second scenario, there is a probability $L - (ell_1 + 2L/3) = L/3 - ell_1$ that we have the desired property.
Averaging over $ell_1$, we find that the desired probability is
$$ textstyle int_0^L/3 (L/3 - ell) f_1(ell) dell
quadtextwherequad
f_1(ell) = 1/(L/2) = 2/L text is the pdf of $ell_1$. $$
Note that this integral is equal to
$$ textstyle (2/L) int_0^L/3 ell' d ell' = (2/L) cdot tfrac12 (L/3)^2 = L/9. $$

share|cite|improve this answer

answered Mar 21 at 22:20

Sam TSam T

3,9401031

$endgroup$

add a comment | 

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3

$begingroup$

Does this make sense to you? I think this aligns with what Ethan has commented.

share|cite|improve this answer

answered Mar 21 at 14:42

J. WangJ. Wang

1916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah. I got it.
  $endgroup$
  – ChakSayantan
  Mar 21 at 14:47
  

  
  
  
  

add a comment | 

3

$begingroup$

Does this make sense to you? I think this aligns with what Ethan has commented.

share|cite|improve this answer

answered Mar 21 at 14:42

J. WangJ. Wang

1916

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yeah. I got it.
  $endgroup$
  – ChakSayantan
  Mar 21 at 14:47
  

  
  
  
  

add a comment | 

$begingroup$

Does this make sense to you? I think this aligns with what Ethan has commented.

share|cite|improve this answer

answered Mar 21 at 14:42

J. WangJ. Wang

1916

$endgroup$

share|cite|improve this answer

answered Mar 21 at 14:42

J. WangJ. Wang

1916

answered Mar 21 at 14:42

J. WangJ. Wang

1916

answered Mar 21 at 14:42

J. WangJ. Wang

1916

0

$begingroup$

A picture, as provided by J. Wang, is very useful for intuition (and has my +1!), but it does need to be made into a rigorous proof. My suggestion is to try conditioning on the first point, $ell_1$. Once you know this, you can ask "Which options for $ell_2$ give the required condition?"---this clearly depends on $ell_1$.

• If $ell_1 > L/3$, then clearly no $ell_2$ will suffice.


• If $ell_1 in [0,L/3]$, then any $ell_2 in [ell_1+2L/3,L]$ will suffice.

Given $ell_1$, and that we land in the second scenario, there is a probability $L - (ell_1 + 2L/3) = L/3 - ell_1$ that we have the desired property.
Averaging over $ell_1$, we find that the desired probability is
$$ textstyle int_0^L/3 (L/3 - ell) f_1(ell) dell
quadtextwherequad
f_1(ell) = 1/(L/2) = 2/L text is the pdf of $ell_1$. $$
Note that this integral is equal to
$$ textstyle (2/L) int_0^L/3 ell' d ell' = (2/L) cdot tfrac12 (L/3)^2 = L/9. $$

share|cite|improve this answer

answered Mar 21 at 22:20

Sam TSam T

3,9401031

$endgroup$

add a comment | 

0

$begingroup$

A picture, as provided by J. Wang, is very useful for intuition (and has my +1!), but it does need to be made into a rigorous proof. My suggestion is to try conditioning on the first point, $ell_1$. Once you know this, you can ask "Which options for $ell_2$ give the required condition?"---this clearly depends on $ell_1$.

• If $ell_1 > L/3$, then clearly no $ell_2$ will suffice.


• If $ell_1 in [0,L/3]$, then any $ell_2 in [ell_1+2L/3,L]$ will suffice.

Given $ell_1$, and that we land in the second scenario, there is a probability $L - (ell_1 + 2L/3) = L/3 - ell_1$ that we have the desired property.
Averaging over $ell_1$, we find that the desired probability is
$$ textstyle int_0^L/3 (L/3 - ell) f_1(ell) dell
quadtextwherequad
f_1(ell) = 1/(L/2) = 2/L text is the pdf of $ell_1$. $$
Note that this integral is equal to
$$ textstyle (2/L) int_0^L/3 ell' d ell' = (2/L) cdot tfrac12 (L/3)^2 = L/9. $$

share|cite|improve this answer

answered Mar 21 at 22:20

Sam TSam T

3,9401031

$endgroup$

add a comment | 

$begingroup$

A picture, as provided by J. Wang, is very useful for intuition (and has my +1!), but it does need to be made into a rigorous proof. My suggestion is to try conditioning on the first point, $ell_1$. Once you know this, you can ask "Which options for $ell_2$ give the required condition?"---this clearly depends on $ell_1$.

• If $ell_1 > L/3$, then clearly no $ell_2$ will suffice.


• If $ell_1 in [0,L/3]$, then any $ell_2 in [ell_1+2L/3,L]$ will suffice.

Given $ell_1$, and that we land in the second scenario, there is a probability $L - (ell_1 + 2L/3) = L/3 - ell_1$ that we have the desired property.
Averaging over $ell_1$, we find that the desired probability is
$$ textstyle int_0^L/3 (L/3 - ell) f_1(ell) dell
quadtextwherequad
f_1(ell) = 1/(L/2) = 2/L text is the pdf of $ell_1$. $$
Note that this integral is equal to
$$ textstyle (2/L) int_0^L/3 ell' d ell' = (2/L) cdot tfrac12 (L/3)^2 = L/9. $$

share|cite|improve this answer

answered Mar 21 at 22:20

Sam TSam T

3,9401031

$endgroup$

share|cite|improve this answer

answered Mar 21 at 22:20

Sam TSam T

3,9401031

answered Mar 21 at 22:20

Sam TSam T

3,9401031

answered Mar 21 at 22:20

Sam TSam T

3,9401031