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Probability of a point $X$ chosen at random on a line segment $AB$ followed by some conditions.
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A point $X$ chosen at random on a line segment $AB$. Show that the
probability that the ratio of the length of the shorter segment to
that of the larger segment is less than $frac13$ is $frac12$
My input
I know the working of this problem and can arrive at exact answer easily with the following steps. But I am not aware of the reasoning behind a few steps. I just know the procedure. So help me out here.
$AB=l$ ; Assuming that $xsim U(0,l)$
Case $1$.
When the point is chosen lies to the left of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (shorter).
Case $2$.
When the point is chosen lies to the right of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (larger).
$Pbigg(dfracxl-x < dfrac13 bigg) cup Pbigg(dfracl-xx < dfrac13 bigg) $
$P(x<dfracl4)+ P(x>dfrac3l4)$
Well solving this we get $dfrac12$
My doubts are:
In the beginning, I wrote $xsim U(0,l)$ I am not getting why are we doing this in my head. I have this crammed that all the probability questions related to distances are done with the help of uniform distribution and that's why I just wrote that. I want to know the exact reasoning why we did that. Also in some questions related to stick problem(like breaking stick from middle etc), we assumed that $X sim U(0,1)$. Why did we do that? Stick length can also be $2$ or $20$ may be. Someone clear this to me, please. Sorry for bad English.
probability probability-theory probability-distributions
edited yesterday
Vobo
2,3961119
asked 2 days ago
Damn1o1
56612
add a comment |Â
up vote
2
down vote
favorite
A point $X$ chosen at random on a line segment $AB$. Show that the
probability that the ratio of the length of the shorter segment to
that of the larger segment is less than $frac13$ is $frac12$
My input
I know the working of this problem and can arrive at exact answer easily with the following steps. But I am not aware of the reasoning behind a few steps. I just know the procedure. So help me out here.
$AB=l$ ; Assuming that $xsim U(0,l)$
Case $1$.
When the point is chosen lies to the left of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (shorter).
Case $2$.
When the point is chosen lies to the right of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (larger).
$Pbigg(dfracxl-x < dfrac13 bigg) cup Pbigg(dfracl-xx < dfrac13 bigg) $
$P(x<dfracl4)+ P(x>dfrac3l4)$
Well solving this we get $dfrac12$
My doubts are:
In the beginning, I wrote $xsim U(0,l)$ I am not getting why are we doing this in my head. I have this crammed that all the probability questions related to distances are done with the help of uniform distribution and that's why I just wrote that. I want to know the exact reasoning why we did that. Also in some questions related to stick problem(like breaking stick from middle etc), we assumed that $X sim U(0,1)$. Why did we do that? Stick length can also be $2$ or $20$ may be. Someone clear this to me, please. Sorry for bad English.
probability probability-theory probability-distributions
edited yesterday
Vobo
2,3961119
asked 2 days ago
Damn1o1
56612
The statement "chosen at random" without any further precision implies uniform probability. So you can either take distance from the left ($mathbb U(0,l)$) or proportion of the total length ($mathbb U(0,1)$).
– Nicolas FRANCOIS
2 days ago
@NicolasFRANCOIS I didn't get "proportion of total length $U(0,1)$" part.
– Damn1o1
2 days ago
If $X$ is the chosen point, and $[AB]$ the segment representing the stick, you can chose $x=AX$ (distance) or $x=fracAXAB$ (proportion of the distance to the total length) as your random variable.
– Nicolas FRANCOIS
2 days ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A point $X$ chosen at random on a line segment $AB$. Show that the
probability that the ratio of the length of the shorter segment to
that of the larger segment is less than $frac13$ is $frac12$
My input
I know the working of this problem and can arrive at exact answer easily with the following steps. But I am not aware of the reasoning behind a few steps. I just know the procedure. So help me out here.
$AB=l$ ; Assuming that $xsim U(0,l)$
Case $1$.
When the point is chosen lies to the left of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (shorter).
Case $2$.
When the point is chosen lies to the right of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (larger).
$Pbigg(dfracxl-x < dfrac13 bigg) cup Pbigg(dfracl-xx < dfrac13 bigg) $
$P(x<dfracl4)+ P(x>dfrac3l4)$
Well solving this we get $dfrac12$
My doubts are:
In the beginning, I wrote $xsim U(0,l)$ I am not getting why are we doing this in my head. I have this crammed that all the probability questions related to distances are done with the help of uniform distribution and that's why I just wrote that. I want to know the exact reasoning why we did that. Also in some questions related to stick problem(like breaking stick from middle etc), we assumed that $X sim U(0,1)$. Why did we do that? Stick length can also be $2$ or $20$ may be. Someone clear this to me, please. Sorry for bad English.
probability probability-theory probability-distributions
edited yesterday
Vobo
2,3961119
asked 2 days ago
Damn1o1
56612
A point $X$ chosen at random on a line segment $AB$. Show that the
probability that the ratio of the length of the shorter segment to
that of the larger segment is less than $frac13$ is $frac12$
My input
I know the working of this problem and can arrive at exact answer easily with the following steps. But I am not aware of the reasoning behind a few steps. I just know the procedure. So help me out here.
$AB=l$ ; Assuming that $xsim U(0,l)$
Case $1$.
When the point is chosen lies to the left of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (shorter).
Case $2$.
When the point is chosen lies to the right of the midpoint we have:
$x$ be the length of $AX$ and $l-x$ be the length of $XB$ (larger).
$Pbigg(dfracxl-x < dfrac13 bigg) cup Pbigg(dfracl-xx < dfrac13 bigg) $
$P(x<dfracl4)+ P(x>dfrac3l4)$
Well solving this we get $dfrac12$
My doubts are:
In the beginning, I wrote $xsim U(0,l)$ I am not getting why are we doing this in my head. I have this crammed that all the probability questions related to distances are done with the help of uniform distribution and that's why I just wrote that. I want to know the exact reasoning why we did that. Also in some questions related to stick problem(like breaking stick from middle etc), we assumed that $X sim U(0,1)$. Why did we do that? Stick length can also be $2$ or $20$ may be. Someone clear this to me, please. Sorry for bad English.
probability probability-theory probability-distributions
edited yesterday
Vobo
2,3961119
asked 2 days ago
Damn1o1
56612
edited yesterday
Vobo
2,3961119
edited yesterday
Vobo
2,3961119
edited yesterday
Vobo
2,3961119
2,3961119
asked 2 days ago
Damn1o1
56612
asked 2 days ago
Damn1o1
56612
asked 2 days ago
Damn1o1
56612
56612
The statement "chosen at random" without any further precision implies uniform probability. So you can either take distance from the left ($mathbb U(0,l)$) or proportion of the total length ($mathbb U(0,1)$).
– Nicolas FRANCOIS
2 days ago
@NicolasFRANCOIS I didn't get "proportion of total length $U(0,1)$" part.
– Damn1o1
2 days ago
If $X$ is the chosen point, and $[AB]$ the segment representing the stick, you can chose $x=AX$ (distance) or $x=fracAXAB$ (proportion of the distance to the total length) as your random variable.
– Nicolas FRANCOIS
2 days ago
add a comment |Â
The statement "chosen at random" without any further precision implies uniform probability. So you can either take distance from the left ($mathbb U(0,l)$) or proportion of the total length ($mathbb U(0,1)$).
– Nicolas FRANCOIS
2 days ago
@NicolasFRANCOIS I didn't get "proportion of total length $U(0,1)$" part.
– Damn1o1
2 days ago
If $X$ is the chosen point, and $[AB]$ the segment representing the stick, you can chose $x=AX$ (distance) or $x=fracAXAB$ (proportion of the distance to the total length) as your random variable.
– Nicolas FRANCOIS
2 days ago
The statement "chosen at random" without any further precision implies uniform probability. So you can either take distance from the left ($mathbb U(0,l)$) or proportion of the total length ($mathbb U(0,1)$).
– Nicolas FRANCOIS
2 days ago
The statement "chosen at random" without any further precision implies uniform probability. So you can either take distance from the left ($mathbb U(0,l)$) or proportion of the total length ($mathbb U(0,1)$).
– Nicolas FRANCOIS
2 days ago
@NicolasFRANCOIS I didn't get "proportion of total length $U(0,1)$" part.
– Damn1o1
2 days ago
@NicolasFRANCOIS I didn't get "proportion of total length $U(0,1)$" part.
– Damn1o1
2 days ago
If $X$ is the chosen point, and $[AB]$ the segment representing the stick, you can chose $x=AX$ (distance) or $x=fracAXAB$ (proportion of the distance to the total length) as your random variable.
– Nicolas FRANCOIS
2 days ago
If $X$ is the chosen point, and $[AB]$ the segment representing the stick, you can chose $x=AX$ (distance) or $x=fracAXAB$ (proportion of the distance to the total length) as your random variable.
– Nicolas FRANCOIS
2 days ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
When you pick a point at random, you give every point on the stick the same probability, hence the distribution is uniform.
As you noticed in probability problems with the stick, the actual length of the stick is irrelevant, so you may assume $l=1$ and work with $U(0,1)$ instead of $U(0,l$ or $U(a,b)$. The final result will be the same.
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
When you pick a point at random, you give every point on the stick the same probability, hence the distribution is uniform.
As you noticed in probability problems with the stick, the actual length of the stick is irrelevant, so you may assume $l=1$ and work with $U(0,1)$ instead of $U(0,l$ or $U(a,b)$. The final result will be the same.
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
add a comment |Â
up vote
1
down vote
When you pick a point at random, you give every point on the stick the same probability, hence the distribution is uniform.
As you noticed in probability problems with the stick, the actual length of the stick is irrelevant, so you may assume $l=1$ and work with $U(0,1)$ instead of $U(0,l$ or $U(a,b)$. The final result will be the same.
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
add a comment |Â
up vote
1
down vote
up vote
1
down vote
When you pick a point at random, you give every point on the stick the same probability, hence the distribution is uniform.
As you noticed in probability problems with the stick, the actual length of the stick is irrelevant, so you may assume $l=1$ and work with $U(0,1)$ instead of $U(0,l$ or $U(a,b)$. The final result will be the same.
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
When you pick a point at random, you give every point on the stick the same probability, hence the distribution is uniform.
As you noticed in probability problems with the stick, the actual length of the stick is irrelevant, so you may assume $l=1$ and work with $U(0,1)$ instead of $U(0,l$ or $U(a,b)$. The final result will be the same.
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
answered 2 days ago
Mohammad Riazi-Kermani
27k41850
27k41850
add a comment |Â
add a comment |Â
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The statement "chosen at random" without any further precision implies uniform probability. So you can either take distance from the left ($mathbb U(0,l)$) or proportion of the total length ($mathbb U(0,1)$).
– Nicolas FRANCOIS
2 days ago
@NicolasFRANCOIS I didn't get "proportion of total length $U(0,1)$" part.
– Damn1o1
2 days ago
If $X$ is the chosen point, and $[AB]$ the segment representing the stick, you can chose $x=AX$ (distance) or $x=fracAXAB$ (proportion of the distance to the total length) as your random variable.
– Nicolas FRANCOIS
2 days ago