Pick $2$ numbers from $[-1,1]$,what is the probability that their sum is greater than $1$? The 2019 Stack Overflow Developer Survey Results Are InChoosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?Probability that 3 number chosen from set is greater than other 3 numbersIf $X$ ~ $U[0, 4]$ and $Y$~$[0, 7]$ find the probability X value is greater than Y valueWhat is the probability that the product of $20$ random numbers between $1$ and $2$ is greater than $10000$?What is the probability that $triangle ABP$ has a greater area than each of $triangle ACP$ and $triangle BCP$?What is the probability that two numbers between 1 and 10 picked at random sum to a number greater than 5?What is the probability that a point chosen randomly from inside an equilateral triangle is closer to the center than to any of the edges?Probability of two uniform random numbers being more than $frac12$ apartIf I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20?Sum of two independent random variable (convolution)
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Pick $2$ numbers from $[-1,1]$,what is the probability that their sum is greater than $1$?
The 2019 Stack Overflow Developer Survey Results Are InChoosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?Probability that 3 number chosen from set is greater than other 3 numbersIf $X$ ~ $U[0, 4]$ and $Y$~$[0, 7]$ find the probability X value is greater than Y valueWhat is the probability that the product of $20$ random numbers between $1$ and $2$ is greater than $10000$?What is the probability that $triangle ABP$ has a greater area than each of $triangle ACP$ and $triangle BCP$?What is the probability that two numbers between 1 and 10 picked at random sum to a number greater than 5?What is the probability that a point chosen randomly from inside an equilateral triangle is closer to the center than to any of the edges?Probability of two uniform random numbers being more than $frac12$ apartIf I select 9 random numbers from 1 to 10, what is the probability that their sum is less than 20?Sum of two independent random variable (convolution)
$begingroup$
Pick 2 numbers from $[-1,1]$, what is the probability that their sum is greater than 1?
It is equal to the probability that the sum of 2 uniform random variables on $[-1,1]$ is greater than 1?
so far,
I got $f(x) = 1/2$ $(-1 < x < 1)$ and $f(y) = 1/2$ $(-1 < y < 1)$, I need to calculate $P(X + Y > 1)$.
I plot the picture of the above convolution,
it is a triangle with vertices on $(-2,0),(2,0),(0,1/2)$.
So $P(X + Y > 1)$ is the area to the right side of $x = 1$, which is $1/2 * 1/4 * 1 = 1/8$, is this correct?
Update:Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?
probability convolution uniform-distribution
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
$endgroup$
add a comment |
$begingroup$
Pick 2 numbers from $[-1,1]$, what is the probability that their sum is greater than 1?
It is equal to the probability that the sum of 2 uniform random variables on $[-1,1]$ is greater than 1?
so far,
I got $f(x) = 1/2$ $(-1 < x < 1)$ and $f(y) = 1/2$ $(-1 < y < 1)$, I need to calculate $P(X + Y > 1)$.
I plot the picture of the above convolution,
it is a triangle with vertices on $(-2,0),(2,0),(0,1/2)$.
So $P(X + Y > 1)$ is the area to the right side of $x = 1$, which is $1/2 * 1/4 * 1 = 1/8$, is this correct?
Update:Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?
probability convolution uniform-distribution
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
$endgroup$
add a comment |
$begingroup$
Pick 2 numbers from $[-1,1]$, what is the probability that their sum is greater than 1?
It is equal to the probability that the sum of 2 uniform random variables on $[-1,1]$ is greater than 1?
so far,
I got $f(x) = 1/2$ $(-1 < x < 1)$ and $f(y) = 1/2$ $(-1 < y < 1)$, I need to calculate $P(X + Y > 1)$.
I plot the picture of the above convolution,
it is a triangle with vertices on $(-2,0),(2,0),(0,1/2)$.
So $P(X + Y > 1)$ is the area to the right side of $x = 1$, which is $1/2 * 1/4 * 1 = 1/8$, is this correct?
Update:Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?
probability convolution uniform-distribution
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
$endgroup$
Pick 2 numbers from $[-1,1]$, what is the probability that their sum is greater than 1?
It is equal to the probability that the sum of 2 uniform random variables on $[-1,1]$ is greater than 1?
so far,
I got $f(x) = 1/2$ $(-1 < x < 1)$ and $f(y) = 1/2$ $(-1 < y < 1)$, I need to calculate $P(X + Y > 1)$.
I plot the picture of the above convolution,
it is a triangle with vertices on $(-2,0),(2,0),(0,1/2)$.
So $P(X + Y > 1)$ is the area to the right side of $x = 1$, which is $1/2 * 1/4 * 1 = 1/8$, is this correct?
Update:Choosing two random numbers in $(0,1)$ what is the probability that sum of them is more than $1$?
probability convolution uniform-distribution
probability convolution uniform-distribution
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
edited Apr 8 at 15:14
evergreenhomeland
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
asked Apr 8 at 8:51
evergreenhomelandevergreenhomeland
998
998
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Yes, that is right. An alternative way to get the same answer is to argue as follows. For the sum to exceed $1$, we need both variables to be positive, so $$P(X+Y>1)=P(X,Y>0)times P(X+Y>1mid X,Y>0).$$
Now $P(X,Y>0)=frac12timesfrac12$. Conditioning on this, gives independent uniform variables on $[0,1]$. If $A,B$ independent uniform on $[0,1]$, $A+B>1$ if and only if $(1-A)+(1-B)<1$, and $(1-A),(1-B)$ are also independent uniform on $[0,1]$. It follows that $P(X+Y>1mid X,Y>0)=frac12$, giving an overall probability of $frac12timesfrac12timesfrac12$.
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
$endgroup$
add a comment |
$begingroup$
I think the quickest way to see this is geometrically. Choosing two points uniformly on $[-1,1]$ is the same as choosing one point uniformly in the box $[-1,1]times[-1,1]subsetmathbbR^2$. Precisely $frac18$ of this square is above the line $x+y=1$.
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
$endgroup$
1
$begingroup$
simply elegant.
$endgroup$
– AKroell
Apr 8 at 10:19
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
Yes, that is right. An alternative way to get the same answer is to argue as follows. For the sum to exceed $1$, we need both variables to be positive, so $$P(X+Y>1)=P(X,Y>0)times P(X+Y>1mid X,Y>0).$$
Now $P(X,Y>0)=frac12timesfrac12$. Conditioning on this, gives independent uniform variables on $[0,1]$. If $A,B$ independent uniform on $[0,1]$, $A+B>1$ if and only if $(1-A)+(1-B)<1$, and $(1-A),(1-B)$ are also independent uniform on $[0,1]$. It follows that $P(X+Y>1mid X,Y>0)=frac12$, giving an overall probability of $frac12timesfrac12timesfrac12$.
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
$endgroup$
add a comment |
$begingroup$
Yes, that is right. An alternative way to get the same answer is to argue as follows. For the sum to exceed $1$, we need both variables to be positive, so $$P(X+Y>1)=P(X,Y>0)times P(X+Y>1mid X,Y>0).$$
Now $P(X,Y>0)=frac12timesfrac12$. Conditioning on this, gives independent uniform variables on $[0,1]$. If $A,B$ independent uniform on $[0,1]$, $A+B>1$ if and only if $(1-A)+(1-B)<1$, and $(1-A),(1-B)$ are also independent uniform on $[0,1]$. It follows that $P(X+Y>1mid X,Y>0)=frac12$, giving an overall probability of $frac12timesfrac12timesfrac12$.
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
$endgroup$
add a comment |
$begingroup$
Yes, that is right. An alternative way to get the same answer is to argue as follows. For the sum to exceed $1$, we need both variables to be positive, so $$P(X+Y>1)=P(X,Y>0)times P(X+Y>1mid X,Y>0).$$
Now $P(X,Y>0)=frac12timesfrac12$. Conditioning on this, gives independent uniform variables on $[0,1]$. If $A,B$ independent uniform on $[0,1]$, $A+B>1$ if and only if $(1-A)+(1-B)<1$, and $(1-A),(1-B)$ are also independent uniform on $[0,1]$. It follows that $P(X+Y>1mid X,Y>0)=frac12$, giving an overall probability of $frac12timesfrac12timesfrac12$.
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
$endgroup$
Yes, that is right. An alternative way to get the same answer is to argue as follows. For the sum to exceed $1$, we need both variables to be positive, so $$P(X+Y>1)=P(X,Y>0)times P(X+Y>1mid X,Y>0).$$
Now $P(X,Y>0)=frac12timesfrac12$. Conditioning on this, gives independent uniform variables on $[0,1]$. If $A,B$ independent uniform on $[0,1]$, $A+B>1$ if and only if $(1-A)+(1-B)<1$, and $(1-A),(1-B)$ are also independent uniform on $[0,1]$. It follows that $P(X+Y>1mid X,Y>0)=frac12$, giving an overall probability of $frac12timesfrac12timesfrac12$.
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
answered Apr 8 at 9:04
Especially LimeEspecially Lime
22.9k23059
22.9k23059
add a comment |
add a comment |
$begingroup$
I think the quickest way to see this is geometrically. Choosing two points uniformly on $[-1,1]$ is the same as choosing one point uniformly in the box $[-1,1]times[-1,1]subsetmathbbR^2$. Precisely $frac18$ of this square is above the line $x+y=1$.
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
$endgroup$
1
$begingroup$
simply elegant.
$endgroup$
– AKroell
Apr 8 at 10:19
add a comment |
$begingroup$
I think the quickest way to see this is geometrically. Choosing two points uniformly on $[-1,1]$ is the same as choosing one point uniformly in the box $[-1,1]times[-1,1]subsetmathbbR^2$. Precisely $frac18$ of this square is above the line $x+y=1$.
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
$endgroup$
1
$begingroup$
simply elegant.
$endgroup$
– AKroell
Apr 8 at 10:19
add a comment |
$begingroup$
I think the quickest way to see this is geometrically. Choosing two points uniformly on $[-1,1]$ is the same as choosing one point uniformly in the box $[-1,1]times[-1,1]subsetmathbbR^2$. Precisely $frac18$ of this square is above the line $x+y=1$.
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
$endgroup$
I think the quickest way to see this is geometrically. Choosing two points uniformly on $[-1,1]$ is the same as choosing one point uniformly in the box $[-1,1]times[-1,1]subsetmathbbR^2$. Precisely $frac18$ of this square is above the line $x+y=1$.
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
answered Apr 8 at 10:11
G Tony JacobsG Tony Jacobs
26k43686
26k43686
1
$begingroup$
simply elegant.
$endgroup$
– AKroell
Apr 8 at 10:19
add a comment |
1
$begingroup$
simply elegant.
$endgroup$
– AKroell
Apr 8 at 10:19
1
1
$begingroup$
simply elegant.
$endgroup$
– AKroell
Apr 8 at 10:19
$begingroup$
simply elegant.
$endgroup$
– AKroell
Apr 8 at 10:19
add a comment |
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