Mixing
3 people distributed at random problem
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Along a road $1$ mile long are $3$ people distributed at random. Find the probability that no $2$ people are less than a distance of d miles apart when $d leq frac12$.
I have read this solution:
Let $x_i$ be the position of the $i^th$ person. I am not sure how to get the following joint density?
$f(x_1,x_2,x_3)=3!$, $0<x_1<x_2<x_3<1$
Is there any other solution to this problem that is more intuitive?
probability
edited Jul 26 at 15:08
Key Flex
4,083423
asked Jul 25 at 20:20
emily
493
add a comment |Â
up vote
0
down vote
favorite
Along a road $1$ mile long are $3$ people distributed at random. Find the probability that no $2$ people are less than a distance of d miles apart when $d leq frac12$.
I have read this solution:
Let $x_i$ be the position of the $i^th$ person. I am not sure how to get the following joint density?
$f(x_1,x_2,x_3)=3!$, $0<x_1<x_2<x_3<1$
Is there any other solution to this problem that is more intuitive?
probability
edited Jul 26 at 15:08
Key Flex
4,083423
asked Jul 25 at 20:20
emily
493
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Along a road $1$ mile long are $3$ people distributed at random. Find the probability that no $2$ people are less than a distance of d miles apart when $d leq frac12$.
I have read this solution:
Let $x_i$ be the position of the $i^th$ person. I am not sure how to get the following joint density?
$f(x_1,x_2,x_3)=3!$, $0<x_1<x_2<x_3<1$
Is there any other solution to this problem that is more intuitive?
probability
edited Jul 26 at 15:08
Key Flex
4,083423
asked Jul 25 at 20:20
emily
493
Along a road $1$ mile long are $3$ people distributed at random. Find the probability that no $2$ people are less than a distance of d miles apart when $d leq frac12$.
I have read this solution:
Let $x_i$ be the position of the $i^th$ person. I am not sure how to get the following joint density?
$f(x_1,x_2,x_3)=3!$, $0<x_1<x_2<x_3<1$
Is there any other solution to this problem that is more intuitive?
probability
edited Jul 26 at 15:08
Key Flex
4,083423
asked Jul 25 at 20:20
emily
493
edited Jul 26 at 15:08
Key Flex
4,083423
edited Jul 26 at 15:08
Key Flex
4,083423
edited Jul 26 at 15:08
Key Flex
4,083423
4,083423
asked Jul 25 at 20:20
emily
493
asked Jul 25 at 20:20
emily
493
asked Jul 25 at 20:20
emily
493
493
add a comment |Â
add a comment |Â
3 Answers
3
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oldest
votes
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0
down vote
What is your solution again? I do not understand. This is a continuous problem, what would 3! represent here? You need to work with the joint probability density function $f: [0,1]^3 rightarrow mathbbR$, $f$ is constant 1.
Hint: compute the probability of the complement event.
It consists of those points $x,y,z$ in all permutations such that $ygeq x+d$ and $zgeq y+d$.
answered Jul 25 at 20:32
A. Pongrácz
1,802115
add a comment |Â
up vote
0
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Consider random variables,$X_i,i= 1,2,3$, representing the positions of the $3$ persons on the road and express the desired probability in terms of the associated order statistics.
Let us assume that â€Âdistributed at random†means that the position of the $3$ people are independent and uniformly distributed over the road. If $X_i$ denotes the position of the $i^th$ person, then the desired probability is
$$P(X_i>X_(i-1)+d,i=2,3)$$Because
$$fX_(1),X_(2),X_(3)(x_1,x_2,x_3)=3!,0<x_1<x_2<x_3<1$$ if follows that $$ZZZ$$$$P(X_(i)>X_(i-1)+d,i=2,3)=fX_(1),X_(2),X_(3)(x_1,x_2,x_3)dx_2dx_2dx_3$$
answered Jul 25 at 20:51
Key Flex
4,083423
add a comment |Â
up vote
0
down vote
This follows from symmetry. In particular, consider all possible sets of orderings of $x_1, x_2$, and $x_3$. (So let $A_1 = x_1 < x_2 < x_3 , A_2 = x_2 < x_1 < x_3 ,...$. How many are there? Let that be $N$. (How many ways can you sort $x_1, x_2$ and $x_3$? That is $N$)
Now, note that the union of all these sets is almost $[0,1]^3$, which has area 1. (You only did not include the cases where any two are equal, but what is the probability of $X_1 = X_2$ if they are i.i.d continuous random variables?). Since you know that these sets have the same volume by the symmetry, the volume of each of these is $frac1N$.
answered Jul 25 at 20:51
E-A
1,8051312
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
What is your solution again? I do not understand. This is a continuous problem, what would 3! represent here? You need to work with the joint probability density function $f: [0,1]^3 rightarrow mathbbR$, $f$ is constant 1.
Hint: compute the probability of the complement event.
It consists of those points $x,y,z$ in all permutations such that $ygeq x+d$ and $zgeq y+d$.
answered Jul 25 at 20:32
A. Pongrácz
1,802115
add a comment |Â
up vote
0
down vote
What is your solution again? I do not understand. This is a continuous problem, what would 3! represent here? You need to work with the joint probability density function $f: [0,1]^3 rightarrow mathbbR$, $f$ is constant 1.
Hint: compute the probability of the complement event.
It consists of those points $x,y,z$ in all permutations such that $ygeq x+d$ and $zgeq y+d$.
answered Jul 25 at 20:32
A. Pongrácz
1,802115
add a comment |Â
up vote
0
down vote
up vote
0
down vote
What is your solution again? I do not understand. This is a continuous problem, what would 3! represent here? You need to work with the joint probability density function $f: [0,1]^3 rightarrow mathbbR$, $f$ is constant 1.
Hint: compute the probability of the complement event.
It consists of those points $x,y,z$ in all permutations such that $ygeq x+d$ and $zgeq y+d$.
answered Jul 25 at 20:32
A. Pongrácz
1,802115
What is your solution again? I do not understand. This is a continuous problem, what would 3! represent here? You need to work with the joint probability density function $f: [0,1]^3 rightarrow mathbbR$, $f$ is constant 1.
Hint: compute the probability of the complement event.
It consists of those points $x,y,z$ in all permutations such that $ygeq x+d$ and $zgeq y+d$.
answered Jul 25 at 20:32
A. Pongrácz
1,802115
answered Jul 25 at 20:32
A. Pongrácz
1,802115
answered Jul 25 at 20:32
A. Pongrácz
1,802115
answered Jul 25 at 20:32
A. Pongrácz
1,802115
1,802115
add a comment |Â
add a comment |Â
up vote
0
down vote
Consider random variables,$X_i,i= 1,2,3$, representing the positions of the $3$ persons on the road and express the desired probability in terms of the associated order statistics.
Let us assume that â€Âdistributed at random†means that the position of the $3$ people are independent and uniformly distributed over the road. If $X_i$ denotes the position of the $i^th$ person, then the desired probability is
$$P(X_i>X_(i-1)+d,i=2,3)$$Because
$$fX_(1),X_(2),X_(3)(x_1,x_2,x_3)=3!,0<x_1<x_2<x_3<1$$ if follows that $$ZZZ$$$$P(X_(i)>X_(i-1)+d,i=2,3)=fX_(1),X_(2),X_(3)(x_1,x_2,x_3)dx_2dx_2dx_3$$
answered Jul 25 at 20:51
Key Flex
4,083423
add a comment |Â
up vote
0
down vote
Consider random variables,$X_i,i= 1,2,3$, representing the positions of the $3$ persons on the road and express the desired probability in terms of the associated order statistics.
Let us assume that â€Âdistributed at random†means that the position of the $3$ people are independent and uniformly distributed over the road. If $X_i$ denotes the position of the $i^th$ person, then the desired probability is
$$P(X_i>X_(i-1)+d,i=2,3)$$Because
$$fX_(1),X_(2),X_(3)(x_1,x_2,x_3)=3!,0<x_1<x_2<x_3<1$$ if follows that $$ZZZ$$$$P(X_(i)>X_(i-1)+d,i=2,3)=fX_(1),X_(2),X_(3)(x_1,x_2,x_3)dx_2dx_2dx_3$$
answered Jul 25 at 20:51
Key Flex
4,083423
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider random variables,$X_i,i= 1,2,3$, representing the positions of the $3$ persons on the road and express the desired probability in terms of the associated order statistics.
Let us assume that â€Âdistributed at random†means that the position of the $3$ people are independent and uniformly distributed over the road. If $X_i$ denotes the position of the $i^th$ person, then the desired probability is
$$P(X_i>X_(i-1)+d,i=2,3)$$Because
$$fX_(1),X_(2),X_(3)(x_1,x_2,x_3)=3!,0<x_1<x_2<x_3<1$$ if follows that $$ZZZ$$$$P(X_(i)>X_(i-1)+d,i=2,3)=fX_(1),X_(2),X_(3)(x_1,x_2,x_3)dx_2dx_2dx_3$$
answered Jul 25 at 20:51
Key Flex
4,083423
Consider random variables,$X_i,i= 1,2,3$, representing the positions of the $3$ persons on the road and express the desired probability in terms of the associated order statistics.
Let us assume that â€Âdistributed at random†means that the position of the $3$ people are independent and uniformly distributed over the road. If $X_i$ denotes the position of the $i^th$ person, then the desired probability is
$$P(X_i>X_(i-1)+d,i=2,3)$$Because
$$fX_(1),X_(2),X_(3)(x_1,x_2,x_3)=3!,0<x_1<x_2<x_3<1$$ if follows that $$ZZZ$$$$P(X_(i)>X_(i-1)+d,i=2,3)=fX_(1),X_(2),X_(3)(x_1,x_2,x_3)dx_2dx_2dx_3$$
answered Jul 25 at 20:51
Key Flex
4,083423
answered Jul 25 at 20:51
Key Flex
4,083423
answered Jul 25 at 20:51
Key Flex
4,083423
answered Jul 25 at 20:51
Key Flex
4,083423
4,083423
add a comment |Â
add a comment |Â
up vote
0
down vote
This follows from symmetry. In particular, consider all possible sets of orderings of $x_1, x_2$, and $x_3$. (So let $A_1 = x_1 < x_2 < x_3 , A_2 = x_2 < x_1 < x_3 ,...$. How many are there? Let that be $N$. (How many ways can you sort $x_1, x_2$ and $x_3$? That is $N$)
Now, note that the union of all these sets is almost $[0,1]^3$, which has area 1. (You only did not include the cases where any two are equal, but what is the probability of $X_1 = X_2$ if they are i.i.d continuous random variables?). Since you know that these sets have the same volume by the symmetry, the volume of each of these is $frac1N$.
answered Jul 25 at 20:51
E-A
1,8051312
add a comment |Â
up vote
0
down vote
This follows from symmetry. In particular, consider all possible sets of orderings of $x_1, x_2$, and $x_3$. (So let $A_1 = x_1 < x_2 < x_3 , A_2 = x_2 < x_1 < x_3 ,...$. How many are there? Let that be $N$. (How many ways can you sort $x_1, x_2$ and $x_3$? That is $N$)
Now, note that the union of all these sets is almost $[0,1]^3$, which has area 1. (You only did not include the cases where any two are equal, but what is the probability of $X_1 = X_2$ if they are i.i.d continuous random variables?). Since you know that these sets have the same volume by the symmetry, the volume of each of these is $frac1N$.
answered Jul 25 at 20:51
E-A
1,8051312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This follows from symmetry. In particular, consider all possible sets of orderings of $x_1, x_2$, and $x_3$. (So let $A_1 = x_1 < x_2 < x_3 , A_2 = x_2 < x_1 < x_3 ,...$. How many are there? Let that be $N$. (How many ways can you sort $x_1, x_2$ and $x_3$? That is $N$)
Now, note that the union of all these sets is almost $[0,1]^3$, which has area 1. (You only did not include the cases where any two are equal, but what is the probability of $X_1 = X_2$ if they are i.i.d continuous random variables?). Since you know that these sets have the same volume by the symmetry, the volume of each of these is $frac1N$.
answered Jul 25 at 20:51
E-A
1,8051312
This follows from symmetry. In particular, consider all possible sets of orderings of $x_1, x_2$, and $x_3$. (So let $A_1 = x_1 < x_2 < x_3 , A_2 = x_2 < x_1 < x_3 ,...$. How many are there? Let that be $N$. (How many ways can you sort $x_1, x_2$ and $x_3$? That is $N$)
Now, note that the union of all these sets is almost $[0,1]^3$, which has area 1. (You only did not include the cases where any two are equal, but what is the probability of $X_1 = X_2$ if they are i.i.d continuous random variables?). Since you know that these sets have the same volume by the symmetry, the volume of each of these is $frac1N$.
answered Jul 25 at 20:51
E-A
1,8051312
answered Jul 25 at 20:51
E-A
1,8051312
answered Jul 25 at 20:51
E-A
1,8051312
answered Jul 25 at 20:51
E-A
1,8051312
1,8051312
add a comment |Â
add a comment |Â
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