Choosing 2 Points on A line












3












$begingroup$



Two points are selected randomly on a line of length $L$ so as to be on opposite sides of the midpoint of the line. Find the probability that the distance between them is greater than $L/3$.




I was able to figure out that one point will be uniformly distributed on $(0,L/2)$ and the other on $(L/2,L)$, but I don’t know how to proceed further.






probability probability-distributions random-variables







share|cite|improve this question











$endgroup$

















    3












    $begingroup$



    Two points are selected randomly on a line of length $L$ so as to be on opposite sides of the midpoint of the line. Find the probability that the distance between them is greater than $L/3$.




    I was able to figure out that one point will be uniformly distributed on $(0,L/2)$ and the other on $(L/2,L)$, but I don’t know how to proceed further.






    probability probability-distributions random-variables







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$



      Two points are selected randomly on a line of length $L$ so as to be on opposite sides of the midpoint of the line. Find the probability that the distance between them is greater than $L/3$.




      I was able to figure out that one point will be uniformly distributed on $(0,L/2)$ and the other on $(L/2,L)$, but I don’t know how to proceed further.






      probability probability-distributions random-variables







      share|cite|improve this question











      $endgroup$





      Two points are selected randomly on a line of length $L$ so as to be on opposite sides of the midpoint of the line. Find the probability that the distance between them is greater than $L/3$.




      I was able to figure out that one point will be uniformly distributed on $(0,L/2)$ and the other on $(L/2,L)$, but I don’t know how to proceed further.






      probability probability-distributions random-variables




      probability probability-distributions random-variables






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 3 '18 at 12:18









      greedoid

      38.7k114797




      38.7k114797










      asked Dec 2 '18 at 11:30









      user601297user601297

      1346




      1346






















          3 Answers
          3






          active

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          0












          $begingroup$

          Analytic solution. Let $x$ and $y$ be the points in $[0,L]$ such that $0le xle frac L2le yle L$. The PDF is $f(x,y)=frac{4}{L^2}$. We need to find $Pleft(y-xge frac L3right)$.



          Note that $y-xge frac L3 Rightarrow yge x+frac L3ge frac L2 Rightarrow xge frac L6$. Hence, we can divide $x$ into two intervals:
          $$begin{cases}0le xle frac L6\ frac L2le yle Lend{cases}
          text{or} begin{cases}frac L6le xle frac L2\ x+frac L3le yle Lend{cases}$$
          The required probability is:
          $$begin{align}Pleft(y-xge frac L3right)&=int_0^{L/6} int_{L/2}^{L} frac 4{L^2} dydx+int_{L/6}^{L/2} int_{x+L/3}^{L} frac 4{L^2} dydx=\
          &=frac13+frac49=\
          &=frac79.end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @DavidK, thank you, corrected.
            $endgroup$
            – farruhota
            Dec 3 '18 at 17:05



















          3












          $begingroup$

          WLOG we can assume $L=6$. Let $x$ be coordinate of the left point and $y$ of the right.



          Then the sample space is $$Omega = {(x,y);;xin [0,3];;yin [3,6]}$$



          and the event is $$A = {(x,y)inOmega;; ygeq x+2}$$



          so we have a picture:enter image description here



          Now $$P(A) = {Area (A)over Area(Omega)} = {7over 9}$$



          Edit: Yes I missed 3 left squares.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow, this is a really interesting and different approach, thanks man, although the answer given is 7/9, so maybe you’re a little off?
            $endgroup$
            – user601297
            Dec 3 '18 at 1:20





















          1












          $begingroup$

          Call the positions of the points on the line $X_1$ and $X_2$. You have three cases:




          • Case 1: $X_1in [0,L/6]$, in which case the condition is respected. Add the probability of that to your total.

          • Case 2: $X_2in [5L/6,L]$, in which case the condition is also respected. Add the probability of that to your total (taking care to subtract the intersection of "Case 1" with "Case 2" so you don't double count it).

          • Case 3: All other cases. Here I'd think of $X_1$ having a specific value, then find the probability that $X_2$ is placed so that the condition is respected as a function of $X_1$. To make this easy (so you don't have to normalize your probability later), think of the your line being placed in a coordinate system so that $L/6$ is on the point $-1$, $L/2$ is on $0$, and $5L/6$ is on $1$. Finally, integrate over all the values $X_1$ can take. Multiply by the probability that we are in "Case 3", and add to your total.


          I find the probability of success to be 7/9 with this method. Do you?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I don’t really get it completely, especially the third case.
            $endgroup$
            – user601297
            Dec 3 '18 at 1:26










          • $begingroup$
            Looking at @greedoid 's solution you can see graphically the cases I'm counting : case 1 are the three squares on the left, case 2 the three squares on the top (don't double count the top-left corner!) and case 3 is the green triangle that's left over.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:05












          • $begingroup$
            Concerning case 3, I'm saying that you should place $L$ in whichever coordinate system makes your life easy. In this coordinate system $X_1$ is the position of the first point, $a$ and $b$ the minimum and maximum values it can take on in case 3. Evaluate $int^b_a p_S(X_1) dX_1$ where $p_S$ is the probability of success for a fixed value of $X_1$. Make sure you normalize, i.e. $int^b_a 1 dX_1$ should equal $1$.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:15












          • $begingroup$
            ok i got it now, thanks
            $endgroup$
            – user601297
            Dec 3 '18 at 10:07










          • $begingroup$
            If you've already figured out the probability that the distance is greater than half the total length of the segment, you can apply that fact to the segment between $L/6$ and $5L/6,$ multiply by the probability that both points are in that segment, and that's case 3.
            $endgroup$
            – David K
            Dec 3 '18 at 13:56











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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Analytic solution. Let $x$ and $y$ be the points in $[0,L]$ such that $0le xle frac L2le yle L$. The PDF is $f(x,y)=frac{4}{L^2}$. We need to find $Pleft(y-xge frac L3right)$.



          Note that $y-xge frac L3 Rightarrow yge x+frac L3ge frac L2 Rightarrow xge frac L6$. Hence, we can divide $x$ into two intervals:
          $$begin{cases}0le xle frac L6\ frac L2le yle Lend{cases}
          text{or} begin{cases}frac L6le xle frac L2\ x+frac L3le yle Lend{cases}$$
          The required probability is:
          $$begin{align}Pleft(y-xge frac L3right)&=int_0^{L/6} int_{L/2}^{L} frac 4{L^2} dydx+int_{L/6}^{L/2} int_{x+L/3}^{L} frac 4{L^2} dydx=\
          &=frac13+frac49=\
          &=frac79.end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @DavidK, thank you, corrected.
            $endgroup$
            – farruhota
            Dec 3 '18 at 17:05
















          0












          $begingroup$

          Analytic solution. Let $x$ and $y$ be the points in $[0,L]$ such that $0le xle frac L2le yle L$. The PDF is $f(x,y)=frac{4}{L^2}$. We need to find $Pleft(y-xge frac L3right)$.



          Note that $y-xge frac L3 Rightarrow yge x+frac L3ge frac L2 Rightarrow xge frac L6$. Hence, we can divide $x$ into two intervals:
          $$begin{cases}0le xle frac L6\ frac L2le yle Lend{cases}
          text{or} begin{cases}frac L6le xle frac L2\ x+frac L3le yle Lend{cases}$$
          The required probability is:
          $$begin{align}Pleft(y-xge frac L3right)&=int_0^{L/6} int_{L/2}^{L} frac 4{L^2} dydx+int_{L/6}^{L/2} int_{x+L/3}^{L} frac 4{L^2} dydx=\
          &=frac13+frac49=\
          &=frac79.end{align}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @DavidK, thank you, corrected.
            $endgroup$
            – farruhota
            Dec 3 '18 at 17:05














          0












          0








          0





          $begingroup$

          Analytic solution. Let $x$ and $y$ be the points in $[0,L]$ such that $0le xle frac L2le yle L$. The PDF is $f(x,y)=frac{4}{L^2}$. We need to find $Pleft(y-xge frac L3right)$.



          Note that $y-xge frac L3 Rightarrow yge x+frac L3ge frac L2 Rightarrow xge frac L6$. Hence, we can divide $x$ into two intervals:
          $$begin{cases}0le xle frac L6\ frac L2le yle Lend{cases}
          text{or} begin{cases}frac L6le xle frac L2\ x+frac L3le yle Lend{cases}$$
          The required probability is:
          $$begin{align}Pleft(y-xge frac L3right)&=int_0^{L/6} int_{L/2}^{L} frac 4{L^2} dydx+int_{L/6}^{L/2} int_{x+L/3}^{L} frac 4{L^2} dydx=\
          &=frac13+frac49=\
          &=frac79.end{align}$$






          share|cite|improve this answer











          $endgroup$



          Analytic solution. Let $x$ and $y$ be the points in $[0,L]$ such that $0le xle frac L2le yle L$. The PDF is $f(x,y)=frac{4}{L^2}$. We need to find $Pleft(y-xge frac L3right)$.



          Note that $y-xge frac L3 Rightarrow yge x+frac L3ge frac L2 Rightarrow xge frac L6$. Hence, we can divide $x$ into two intervals:
          $$begin{cases}0le xle frac L6\ frac L2le yle Lend{cases}
          text{or} begin{cases}frac L6le xle frac L2\ x+frac L3le yle Lend{cases}$$
          The required probability is:
          $$begin{align}Pleft(y-xge frac L3right)&=int_0^{L/6} int_{L/2}^{L} frac 4{L^2} dydx+int_{L/6}^{L/2} int_{x+L/3}^{L} frac 4{L^2} dydx=\
          &=frac13+frac49=\
          &=frac79.end{align}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 17:04

























          answered Dec 2 '18 at 14:23









          farruhotafarruhota

          19.7k2738




          19.7k2738












          • $begingroup$
            @DavidK, thank you, corrected.
            $endgroup$
            – farruhota
            Dec 3 '18 at 17:05


















          • $begingroup$
            @DavidK, thank you, corrected.
            $endgroup$
            – farruhota
            Dec 3 '18 at 17:05
















          $begingroup$
          @DavidK, thank you, corrected.
          $endgroup$
          – farruhota
          Dec 3 '18 at 17:05




          $begingroup$
          @DavidK, thank you, corrected.
          $endgroup$
          – farruhota
          Dec 3 '18 at 17:05











          3












          $begingroup$

          WLOG we can assume $L=6$. Let $x$ be coordinate of the left point and $y$ of the right.



          Then the sample space is $$Omega = {(x,y);;xin [0,3];;yin [3,6]}$$



          and the event is $$A = {(x,y)inOmega;; ygeq x+2}$$



          so we have a picture:enter image description here



          Now $$P(A) = {Area (A)over Area(Omega)} = {7over 9}$$



          Edit: Yes I missed 3 left squares.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow, this is a really interesting and different approach, thanks man, although the answer given is 7/9, so maybe you’re a little off?
            $endgroup$
            – user601297
            Dec 3 '18 at 1:20


















          3












          $begingroup$

          WLOG we can assume $L=6$. Let $x$ be coordinate of the left point and $y$ of the right.



          Then the sample space is $$Omega = {(x,y);;xin [0,3];;yin [3,6]}$$



          and the event is $$A = {(x,y)inOmega;; ygeq x+2}$$



          so we have a picture:enter image description here



          Now $$P(A) = {Area (A)over Area(Omega)} = {7over 9}$$



          Edit: Yes I missed 3 left squares.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Wow, this is a really interesting and different approach, thanks man, although the answer given is 7/9, so maybe you’re a little off?
            $endgroup$
            – user601297
            Dec 3 '18 at 1:20
















          3












          3








          3





          $begingroup$

          WLOG we can assume $L=6$. Let $x$ be coordinate of the left point and $y$ of the right.



          Then the sample space is $$Omega = {(x,y);;xin [0,3];;yin [3,6]}$$



          and the event is $$A = {(x,y)inOmega;; ygeq x+2}$$



          so we have a picture:enter image description here



          Now $$P(A) = {Area (A)over Area(Omega)} = {7over 9}$$



          Edit: Yes I missed 3 left squares.






          share|cite|improve this answer











          $endgroup$



          WLOG we can assume $L=6$. Let $x$ be coordinate of the left point and $y$ of the right.



          Then the sample space is $$Omega = {(x,y);;xin [0,3];;yin [3,6]}$$



          and the event is $$A = {(x,y)inOmega;; ygeq x+2}$$



          so we have a picture:enter image description here



          Now $$P(A) = {Area (A)over Area(Omega)} = {7over 9}$$



          Edit: Yes I missed 3 left squares.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 13:11

























          answered Dec 2 '18 at 13:34









          greedoidgreedoid

          38.7k114797




          38.7k114797












          • $begingroup$
            Wow, this is a really interesting and different approach, thanks man, although the answer given is 7/9, so maybe you’re a little off?
            $endgroup$
            – user601297
            Dec 3 '18 at 1:20




















          • $begingroup$
            Wow, this is a really interesting and different approach, thanks man, although the answer given is 7/9, so maybe you’re a little off?
            $endgroup$
            – user601297
            Dec 3 '18 at 1:20


















          $begingroup$
          Wow, this is a really interesting and different approach, thanks man, although the answer given is 7/9, so maybe you’re a little off?
          $endgroup$
          – user601297
          Dec 3 '18 at 1:20






          $begingroup$
          Wow, this is a really interesting and different approach, thanks man, although the answer given is 7/9, so maybe you’re a little off?
          $endgroup$
          – user601297
          Dec 3 '18 at 1:20













          1












          $begingroup$

          Call the positions of the points on the line $X_1$ and $X_2$. You have three cases:




          • Case 1: $X_1in [0,L/6]$, in which case the condition is respected. Add the probability of that to your total.

          • Case 2: $X_2in [5L/6,L]$, in which case the condition is also respected. Add the probability of that to your total (taking care to subtract the intersection of "Case 1" with "Case 2" so you don't double count it).

          • Case 3: All other cases. Here I'd think of $X_1$ having a specific value, then find the probability that $X_2$ is placed so that the condition is respected as a function of $X_1$. To make this easy (so you don't have to normalize your probability later), think of the your line being placed in a coordinate system so that $L/6$ is on the point $-1$, $L/2$ is on $0$, and $5L/6$ is on $1$. Finally, integrate over all the values $X_1$ can take. Multiply by the probability that we are in "Case 3", and add to your total.


          I find the probability of success to be 7/9 with this method. Do you?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I don’t really get it completely, especially the third case.
            $endgroup$
            – user601297
            Dec 3 '18 at 1:26










          • $begingroup$
            Looking at @greedoid 's solution you can see graphically the cases I'm counting : case 1 are the three squares on the left, case 2 the three squares on the top (don't double count the top-left corner!) and case 3 is the green triangle that's left over.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:05












          • $begingroup$
            Concerning case 3, I'm saying that you should place $L$ in whichever coordinate system makes your life easy. In this coordinate system $X_1$ is the position of the first point, $a$ and $b$ the minimum and maximum values it can take on in case 3. Evaluate $int^b_a p_S(X_1) dX_1$ where $p_S$ is the probability of success for a fixed value of $X_1$. Make sure you normalize, i.e. $int^b_a 1 dX_1$ should equal $1$.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:15












          • $begingroup$
            ok i got it now, thanks
            $endgroup$
            – user601297
            Dec 3 '18 at 10:07










          • $begingroup$
            If you've already figured out the probability that the distance is greater than half the total length of the segment, you can apply that fact to the segment between $L/6$ and $5L/6,$ multiply by the probability that both points are in that segment, and that's case 3.
            $endgroup$
            – David K
            Dec 3 '18 at 13:56
















          1












          $begingroup$

          Call the positions of the points on the line $X_1$ and $X_2$. You have three cases:




          • Case 1: $X_1in [0,L/6]$, in which case the condition is respected. Add the probability of that to your total.

          • Case 2: $X_2in [5L/6,L]$, in which case the condition is also respected. Add the probability of that to your total (taking care to subtract the intersection of "Case 1" with "Case 2" so you don't double count it).

          • Case 3: All other cases. Here I'd think of $X_1$ having a specific value, then find the probability that $X_2$ is placed so that the condition is respected as a function of $X_1$. To make this easy (so you don't have to normalize your probability later), think of the your line being placed in a coordinate system so that $L/6$ is on the point $-1$, $L/2$ is on $0$, and $5L/6$ is on $1$. Finally, integrate over all the values $X_1$ can take. Multiply by the probability that we are in "Case 3", and add to your total.


          I find the probability of success to be 7/9 with this method. Do you?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I don’t really get it completely, especially the third case.
            $endgroup$
            – user601297
            Dec 3 '18 at 1:26










          • $begingroup$
            Looking at @greedoid 's solution you can see graphically the cases I'm counting : case 1 are the three squares on the left, case 2 the three squares on the top (don't double count the top-left corner!) and case 3 is the green triangle that's left over.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:05












          • $begingroup$
            Concerning case 3, I'm saying that you should place $L$ in whichever coordinate system makes your life easy. In this coordinate system $X_1$ is the position of the first point, $a$ and $b$ the minimum and maximum values it can take on in case 3. Evaluate $int^b_a p_S(X_1) dX_1$ where $p_S$ is the probability of success for a fixed value of $X_1$. Make sure you normalize, i.e. $int^b_a 1 dX_1$ should equal $1$.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:15












          • $begingroup$
            ok i got it now, thanks
            $endgroup$
            – user601297
            Dec 3 '18 at 10:07










          • $begingroup$
            If you've already figured out the probability that the distance is greater than half the total length of the segment, you can apply that fact to the segment between $L/6$ and $5L/6,$ multiply by the probability that both points are in that segment, and that's case 3.
            $endgroup$
            – David K
            Dec 3 '18 at 13:56














          1












          1








          1





          $begingroup$

          Call the positions of the points on the line $X_1$ and $X_2$. You have three cases:




          • Case 1: $X_1in [0,L/6]$, in which case the condition is respected. Add the probability of that to your total.

          • Case 2: $X_2in [5L/6,L]$, in which case the condition is also respected. Add the probability of that to your total (taking care to subtract the intersection of "Case 1" with "Case 2" so you don't double count it).

          • Case 3: All other cases. Here I'd think of $X_1$ having a specific value, then find the probability that $X_2$ is placed so that the condition is respected as a function of $X_1$. To make this easy (so you don't have to normalize your probability later), think of the your line being placed in a coordinate system so that $L/6$ is on the point $-1$, $L/2$ is on $0$, and $5L/6$ is on $1$. Finally, integrate over all the values $X_1$ can take. Multiply by the probability that we are in "Case 3", and add to your total.


          I find the probability of success to be 7/9 with this method. Do you?






          share|cite|improve this answer











          $endgroup$



          Call the positions of the points on the line $X_1$ and $X_2$. You have three cases:




          • Case 1: $X_1in [0,L/6]$, in which case the condition is respected. Add the probability of that to your total.

          • Case 2: $X_2in [5L/6,L]$, in which case the condition is also respected. Add the probability of that to your total (taking care to subtract the intersection of "Case 1" with "Case 2" so you don't double count it).

          • Case 3: All other cases. Here I'd think of $X_1$ having a specific value, then find the probability that $X_2$ is placed so that the condition is respected as a function of $X_1$. To make this easy (so you don't have to normalize your probability later), think of the your line being placed in a coordinate system so that $L/6$ is on the point $-1$, $L/2$ is on $0$, and $5L/6$ is on $1$. Finally, integrate over all the values $X_1$ can take. Multiply by the probability that we are in "Case 3", and add to your total.


          I find the probability of success to be 7/9 with this method. Do you?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 '18 at 13:12

























          answered Dec 2 '18 at 12:19









          JefferyJeffery

          834




          834












          • $begingroup$
            I don’t really get it completely, especially the third case.
            $endgroup$
            – user601297
            Dec 3 '18 at 1:26










          • $begingroup$
            Looking at @greedoid 's solution you can see graphically the cases I'm counting : case 1 are the three squares on the left, case 2 the three squares on the top (don't double count the top-left corner!) and case 3 is the green triangle that's left over.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:05












          • $begingroup$
            Concerning case 3, I'm saying that you should place $L$ in whichever coordinate system makes your life easy. In this coordinate system $X_1$ is the position of the first point, $a$ and $b$ the minimum and maximum values it can take on in case 3. Evaluate $int^b_a p_S(X_1) dX_1$ where $p_S$ is the probability of success for a fixed value of $X_1$. Make sure you normalize, i.e. $int^b_a 1 dX_1$ should equal $1$.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:15












          • $begingroup$
            ok i got it now, thanks
            $endgroup$
            – user601297
            Dec 3 '18 at 10:07










          • $begingroup$
            If you've already figured out the probability that the distance is greater than half the total length of the segment, you can apply that fact to the segment between $L/6$ and $5L/6,$ multiply by the probability that both points are in that segment, and that's case 3.
            $endgroup$
            – David K
            Dec 3 '18 at 13:56


















          • $begingroup$
            I don’t really get it completely, especially the third case.
            $endgroup$
            – user601297
            Dec 3 '18 at 1:26










          • $begingroup$
            Looking at @greedoid 's solution you can see graphically the cases I'm counting : case 1 are the three squares on the left, case 2 the three squares on the top (don't double count the top-left corner!) and case 3 is the green triangle that's left over.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:05












          • $begingroup$
            Concerning case 3, I'm saying that you should place $L$ in whichever coordinate system makes your life easy. In this coordinate system $X_1$ is the position of the first point, $a$ and $b$ the minimum and maximum values it can take on in case 3. Evaluate $int^b_a p_S(X_1) dX_1$ where $p_S$ is the probability of success for a fixed value of $X_1$. Make sure you normalize, i.e. $int^b_a 1 dX_1$ should equal $1$.
            $endgroup$
            – Jeffery
            Dec 3 '18 at 8:15












          • $begingroup$
            ok i got it now, thanks
            $endgroup$
            – user601297
            Dec 3 '18 at 10:07










          • $begingroup$
            If you've already figured out the probability that the distance is greater than half the total length of the segment, you can apply that fact to the segment between $L/6$ and $5L/6,$ multiply by the probability that both points are in that segment, and that's case 3.
            $endgroup$
            – David K
            Dec 3 '18 at 13:56
















          $begingroup$
          I don’t really get it completely, especially the third case.
          $endgroup$
          – user601297
          Dec 3 '18 at 1:26




          $begingroup$
          I don’t really get it completely, especially the third case.
          $endgroup$
          – user601297
          Dec 3 '18 at 1:26












          $begingroup$
          Looking at @greedoid 's solution you can see graphically the cases I'm counting : case 1 are the three squares on the left, case 2 the three squares on the top (don't double count the top-left corner!) and case 3 is the green triangle that's left over.
          $endgroup$
          – Jeffery
          Dec 3 '18 at 8:05






          $begingroup$
          Looking at @greedoid 's solution you can see graphically the cases I'm counting : case 1 are the three squares on the left, case 2 the three squares on the top (don't double count the top-left corner!) and case 3 is the green triangle that's left over.
          $endgroup$
          – Jeffery
          Dec 3 '18 at 8:05














          $begingroup$
          Concerning case 3, I'm saying that you should place $L$ in whichever coordinate system makes your life easy. In this coordinate system $X_1$ is the position of the first point, $a$ and $b$ the minimum and maximum values it can take on in case 3. Evaluate $int^b_a p_S(X_1) dX_1$ where $p_S$ is the probability of success for a fixed value of $X_1$. Make sure you normalize, i.e. $int^b_a 1 dX_1$ should equal $1$.
          $endgroup$
          – Jeffery
          Dec 3 '18 at 8:15






          $begingroup$
          Concerning case 3, I'm saying that you should place $L$ in whichever coordinate system makes your life easy. In this coordinate system $X_1$ is the position of the first point, $a$ and $b$ the minimum and maximum values it can take on in case 3. Evaluate $int^b_a p_S(X_1) dX_1$ where $p_S$ is the probability of success for a fixed value of $X_1$. Make sure you normalize, i.e. $int^b_a 1 dX_1$ should equal $1$.
          $endgroup$
          – Jeffery
          Dec 3 '18 at 8:15














          $begingroup$
          ok i got it now, thanks
          $endgroup$
          – user601297
          Dec 3 '18 at 10:07




          $begingroup$
          ok i got it now, thanks
          $endgroup$
          – user601297
          Dec 3 '18 at 10:07












          $begingroup$
          If you've already figured out the probability that the distance is greater than half the total length of the segment, you can apply that fact to the segment between $L/6$ and $5L/6,$ multiply by the probability that both points are in that segment, and that's case 3.
          $endgroup$
          – David K
          Dec 3 '18 at 13:56




          $begingroup$
          If you've already figured out the probability that the distance is greater than half the total length of the segment, you can apply that fact to the segment between $L/6$ and $5L/6,$ multiply by the probability that both points are in that segment, and that's case 3.
          $endgroup$
          – David K
          Dec 3 '18 at 13:56


















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