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Find the probability that the sum of two randomly chosen positive numbers… followed by some conditions.
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Find the probability that the sum of two randomly chosen positive numbers (both $le 1$) will not exceed $1$ and that their product will be $le frac29$
My input
I just solved a problem in which we choose two points on a unit length line segment and a condition was given that three segments made by these two points must be greater than $frac14$
Similarly I solved this problem as:
Considering $AB$ as a line segment of length $1$ unit.
Same situation is here as we have to choose two numbers(As I took points on line). Also distance cant be negative. Therefor I chose two points such that there sum doesn't succeed $1$ and product is $le frac29$. Two points are chose between $AB$ as $XY$ now length of $AX=x$ ,$XY=y$, And then I just applied condtions as follows:
$x+y<1$ , $XYle frac29$.
I have doubts.
Is this approach is correct?
I don't have its answer behind my book and that's why I didn't calculate shaded region area. I would love to learn any other approach to solve this problem. Sorry if problem seems confusing because of my bad English.
probability probability-theory
asked Jul 31 at 15:34
Damn1o1
57113
 |Â
show 10 more comments
up vote
2
down vote
favorite
Find the probability that the sum of two randomly chosen positive numbers (both $le 1$) will not exceed $1$ and that their product will be $le frac29$
My input
I just solved a problem in which we choose two points on a unit length line segment and a condition was given that three segments made by these two points must be greater than $frac14$
Similarly I solved this problem as:
Considering $AB$ as a line segment of length $1$ unit.
Same situation is here as we have to choose two numbers(As I took points on line). Also distance cant be negative. Therefor I chose two points such that there sum doesn't succeed $1$ and product is $le frac29$. Two points are chose between $AB$ as $XY$ now length of $AX=x$ ,$XY=y$, And then I just applied condtions as follows:
$x+y<1$ , $XYle frac29$.
I have doubts.
Is this approach is correct?
I don't have its answer behind my book and that's why I didn't calculate shaded region area. I would love to learn any other approach to solve this problem. Sorry if problem seems confusing because of my bad English.
probability probability-theory
asked Jul 31 at 15:34
Damn1o1
57113
1
You need to specify the distribution of the two randomly chosen numbers. Are they uniformly distributed?
– Theoretical Economist
Jul 31 at 15:36
Presumably your are using a uniform distribution in which case your approach is a good one.
– copper.hat
Jul 31 at 15:37
@TheoreticalEconomist I assumed that here.
– Damn1o1
Jul 31 at 15:37
1
There is a triangle and a little half banana like area. The probability is the area of the triangle less the half banana area.
– copper.hat
Jul 31 at 15:55
1
I'm confused by what you are asking. You want to compute the yellow area.
– copper.hat
Jul 31 at 15:57
 |Â
show 10 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the probability that the sum of two randomly chosen positive numbers (both $le 1$) will not exceed $1$ and that their product will be $le frac29$
My input
I just solved a problem in which we choose two points on a unit length line segment and a condition was given that three segments made by these two points must be greater than $frac14$
Similarly I solved this problem as:
Considering $AB$ as a line segment of length $1$ unit.
Same situation is here as we have to choose two numbers(As I took points on line). Also distance cant be negative. Therefor I chose two points such that there sum doesn't succeed $1$ and product is $le frac29$. Two points are chose between $AB$ as $XY$ now length of $AX=x$ ,$XY=y$, And then I just applied condtions as follows:
$x+y<1$ , $XYle frac29$.
I have doubts.
Is this approach is correct?
I don't have its answer behind my book and that's why I didn't calculate shaded region area. I would love to learn any other approach to solve this problem. Sorry if problem seems confusing because of my bad English.
probability probability-theory
asked Jul 31 at 15:34
Damn1o1
57113
Find the probability that the sum of two randomly chosen positive numbers (both $le 1$) will not exceed $1$ and that their product will be $le frac29$
My input
I just solved a problem in which we choose two points on a unit length line segment and a condition was given that three segments made by these two points must be greater than $frac14$
Similarly I solved this problem as:
Considering $AB$ as a line segment of length $1$ unit.
Same situation is here as we have to choose two numbers(As I took points on line). Also distance cant be negative. Therefor I chose two points such that there sum doesn't succeed $1$ and product is $le frac29$. Two points are chose between $AB$ as $XY$ now length of $AX=x$ ,$XY=y$, And then I just applied condtions as follows:
$x+y<1$ , $XYle frac29$.
I have doubts.
Is this approach is correct?
I don't have its answer behind my book and that's why I didn't calculate shaded region area. I would love to learn any other approach to solve this problem. Sorry if problem seems confusing because of my bad English.
probability probability-theory
asked Jul 31 at 15:34
Damn1o1
57113
asked Jul 31 at 15:34
Damn1o1
57113
asked Jul 31 at 15:34
Damn1o1
57113
asked Jul 31 at 15:34
Damn1o1
57113
57113
1
You need to specify the distribution of the two randomly chosen numbers. Are they uniformly distributed?
– Theoretical Economist
Jul 31 at 15:36
Presumably your are using a uniform distribution in which case your approach is a good one.
– copper.hat
Jul 31 at 15:37
@TheoreticalEconomist I assumed that here.
– Damn1o1
Jul 31 at 15:37
1
There is a triangle and a little half banana like area. The probability is the area of the triangle less the half banana area.
– copper.hat
Jul 31 at 15:55
1
I'm confused by what you are asking. You want to compute the yellow area.
– copper.hat
Jul 31 at 15:57
 |Â
show 10 more comments
1
You need to specify the distribution of the two randomly chosen numbers. Are they uniformly distributed?
– Theoretical Economist
Jul 31 at 15:36
Presumably your are using a uniform distribution in which case your approach is a good one.
– copper.hat
Jul 31 at 15:37
@TheoreticalEconomist I assumed that here.
– Damn1o1
Jul 31 at 15:37
1
There is a triangle and a little half banana like area. The probability is the area of the triangle less the half banana area.
– copper.hat
Jul 31 at 15:55
1
I'm confused by what you are asking. You want to compute the yellow area.
– copper.hat
Jul 31 at 15:57
1
1
You need to specify the distribution of the two randomly chosen numbers. Are they uniformly distributed?
– Theoretical Economist
Jul 31 at 15:36
You need to specify the distribution of the two randomly chosen numbers. Are they uniformly distributed?
– Theoretical Economist
Jul 31 at 15:36
Presumably your are using a uniform distribution in which case your approach is a good one.
– copper.hat
Jul 31 at 15:37
Presumably your are using a uniform distribution in which case your approach is a good one.
– copper.hat
Jul 31 at 15:37
@TheoreticalEconomist I assumed that here.
– Damn1o1
Jul 31 at 15:37
@TheoreticalEconomist I assumed that here.
– Damn1o1
Jul 31 at 15:37
1
1
There is a triangle and a little half banana like area. The probability is the area of the triangle less the half banana area.
– copper.hat
Jul 31 at 15:55
There is a triangle and a little half banana like area. The probability is the area of the triangle less the half banana area.
– copper.hat
Jul 31 at 15:55
1
1
I'm confused by what you are asking. You want to compute the yellow area.
– copper.hat
Jul 31 at 15:57
I'm confused by what you are asking. You want to compute the yellow area.
– copper.hat
Jul 31 at 15:57
 |Â
show 10 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Comment: You have a path to the answer; this is just a way to confirm
you are on the right track without actually working the problem.
I assume you have independent random variables $X$ and $Y,$ each distributed $mathsfUnif(0, 1).$ And you seek $P(X+Y le 1, XY le 2/9),$ where as usual the comma means intersection.
Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting
the answer by simulation is sometimes called 'Monte Carlo integration'. It is
mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions).
You should use ordinary integration for your answer.]
set.seed(731); m = 10^5; x = runif(m); y = runif(m)
A = (x + y <= 1); B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A); mean(B); mean(A & B) # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086 # aprx P(A) = 1/2
[1] 0.55795 # aprx P(B)
[1] 0.48821 # aprx answer P(A & B)
Because of the order in which I have plotted the 100,000 points below, it is the percentage
of yellow ones that you want, as in your diagram. (About 0.488.)
plot(x, y, pch=".")
points(x[B], y[B], pch=".", col="red") # Read [ ] as "such that"
points(x[A], y[A], pch=".", col="green2")
points(x[A&B], y[A&B], pch=".", col="yellow")
For an exact solution it may be easiest to subtract the area of the green
'nibble' from 1/2. (As @copperhat has suggested.)
answered Aug 1 at 2:30
BruceET
33.1k61440
add a comment |Â
up vote
2
down vote
One way is to compute the area of the triangle ($=1 over 2$) and subtract the area between
the red and blue lines.
The red and blue lines are $y= 2 over 9x$ and $y = 1-x$. Equating and
solving the resulting quadratic gives the bounds of integration
$1 over 2 (1 pm 1 over 3)$. Hence the little area is
$int_1 over 2 (1 - 1 over 3)^1 over 2 (1 + 1 over 3) (1-x-2 over 9x) dx = 1 over 6-2 over 9 ln 2 $.
The probability is then $1 over 3+2 over 9 ln 2 approx 0.487$.
answered Aug 1 at 5:42
copper.hat
122k557155
1
Simulation with a million iterations (too many for a nice figure) agrees with this to three places. (+1) Also clear and to the point.
– BruceET
Aug 2 at 17:16
That's what I calculated. I don't know how I got confused with you in chat. :)
– Damn1o1
Aug 4 at 8:24
add a comment |Â
up vote
1
down vote
Be $X,Y sim U(0,1)$ assuming $X perp Y$ and let $A = X + Y$ and $B=XY$. We’re trying to evaluate:
$$mathrmPleft[A leq 1 cap B leq frac29right]$$
As other answers have already pointed out it is handy to plot the sets. However even if it is correct to say “the probability I’m looking for is the value of that area†given this scenario, I’d like to take a deeper look for why it is that.
Be $D subset mathbbR^2$ the set we’re interested into, which is defined as:
$$D = left(x,y):x+yleq1,xyleqfrac29,xgeq0,ygeq0right$$
Then:
$$
mathrmPleft[A leq 1 cap B leq frac29right]=iint_DmathrmPleft[X = x cap Y = yright],dx,dy
$$
We’re considering every point $(x_0,y_0)$ that satisfies our conditions, that is when $X=x_0$ and $Y=y_0$ are the coordinates of a point inside $D$ at the same time.
Since $Xperp Y$ then:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= 1 cdot 1 \
&= 1
endalign
$$
Now we see why $mathrmPleft[A leq 1 cap B leq frac29right]$ is that area, because:
$$
beginalign
mathrmPleft[A leq 1 cap B leq frac29right]=iint_Ddx,dy=textArea,(D)
endalign
$$
Extra work - while other answers have already found the number you need I’d like to verify what I wrote before. Let’s say that $Xsim U(0,2)$ and $Ysim U(1,4)$. We want to evaluate:
$$
beginalign
mathrmPleft[X+Y leq frac72 cap 2X+Y geq 3right]
endalign
$$
Time to define $D$:
$$D = left(x,y):x+yleqfrac72,2x+ygeq3,0leq x<2,1leq y<4right$$
We’ll plot it using IPython and Matplotlib.
import numpy as np
import matplotlib as mpl
N = 10**6
X = np.random.uniform(0, 2, N)
Y = np.random.uniform(1, 4, N)
A = (X + Y <= 4)
B = (2*X + Y >= 3)
np.mean(A&B) # ~= 0.333
pp.plot(X[A&B], Y[A&B], '.'); pp.show()
We’re going to reference this later to set up the integral. We need to compute now:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= frac12cdotfrac13 \
&= frac16
endalign
$$
Let’s begin:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16iint_Ddx,dy
$$
Looking at $D$ we can easily see a trapezoid from which a triangle has been dropped, then:
$$
iint_Ddx,dy = 2fracfrac12+frac522-frac1cdot 22 = 2
$$
Finally:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16cdot 2 = frac13 simeq 0.333
$$
answered Aug 2 at 1:30
Giulio Scattolin
15619
You should recheck your answer. Other have given the answer so you can confirm it.
– Damn1o1
Aug 4 at 8:25
@Damn1o1 I apologize, you want me to provide the answer using the method I presented?
– Giulio Scattolin
Aug 4 at 9:51
Yes, that will be fine. I want to learn things. A different approach will be better.
– Damn1o1
Aug 4 at 9:53
@Damn1o1 Actually copper.hat's computations are what I would do now, that's way I stopped there. At the moment I don't believe there is another way to get to the answer to your problem without evaluating that area somehow: that area (in this problem) is the probability you're interested to, so any "different approach" will bring you to evaluate that area differently. Instead, you could obtain a different formula starting from the $iint$ but I don't see any benefit since that would be more a "geometric" problem, let's say, more than a "statistic" one.
– Giulio Scattolin
Aug 4 at 10:04
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Comment: You have a path to the answer; this is just a way to confirm
you are on the right track without actually working the problem.
I assume you have independent random variables $X$ and $Y,$ each distributed $mathsfUnif(0, 1).$ And you seek $P(X+Y le 1, XY le 2/9),$ where as usual the comma means intersection.
Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting
the answer by simulation is sometimes called 'Monte Carlo integration'. It is
mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions).
You should use ordinary integration for your answer.]
set.seed(731); m = 10^5; x = runif(m); y = runif(m)
A = (x + y <= 1); B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A); mean(B); mean(A & B) # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086 # aprx P(A) = 1/2
[1] 0.55795 # aprx P(B)
[1] 0.48821 # aprx answer P(A & B)
Because of the order in which I have plotted the 100,000 points below, it is the percentage
of yellow ones that you want, as in your diagram. (About 0.488.)
plot(x, y, pch=".")
points(x[B], y[B], pch=".", col="red") # Read [ ] as "such that"
points(x[A], y[A], pch=".", col="green2")
points(x[A&B], y[A&B], pch=".", col="yellow")
For an exact solution it may be easiest to subtract the area of the green
'nibble' from 1/2. (As @copperhat has suggested.)
answered Aug 1 at 2:30
BruceET
33.1k61440
add a comment |Â
up vote
2
down vote
accepted
Comment: You have a path to the answer; this is just a way to confirm
you are on the right track without actually working the problem.
I assume you have independent random variables $X$ and $Y,$ each distributed $mathsfUnif(0, 1).$ And you seek $P(X+Y le 1, XY le 2/9),$ where as usual the comma means intersection.
Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting
the answer by simulation is sometimes called 'Monte Carlo integration'. It is
mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions).
You should use ordinary integration for your answer.]
set.seed(731); m = 10^5; x = runif(m); y = runif(m)
A = (x + y <= 1); B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A); mean(B); mean(A & B) # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086 # aprx P(A) = 1/2
[1] 0.55795 # aprx P(B)
[1] 0.48821 # aprx answer P(A & B)
Because of the order in which I have plotted the 100,000 points below, it is the percentage
of yellow ones that you want, as in your diagram. (About 0.488.)
plot(x, y, pch=".")
points(x[B], y[B], pch=".", col="red") # Read [ ] as "such that"
points(x[A], y[A], pch=".", col="green2")
points(x[A&B], y[A&B], pch=".", col="yellow")
For an exact solution it may be easiest to subtract the area of the green
'nibble' from 1/2. (As @copperhat has suggested.)
answered Aug 1 at 2:30
BruceET
33.1k61440
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Comment: You have a path to the answer; this is just a way to confirm
you are on the right track without actually working the problem.
I assume you have independent random variables $X$ and $Y,$ each distributed $mathsfUnif(0, 1).$ And you seek $P(X+Y le 1, XY le 2/9),$ where as usual the comma means intersection.
Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting
the answer by simulation is sometimes called 'Monte Carlo integration'. It is
mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions).
You should use ordinary integration for your answer.]
set.seed(731); m = 10^5; x = runif(m); y = runif(m)
A = (x + y <= 1); B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A); mean(B); mean(A & B) # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086 # aprx P(A) = 1/2
[1] 0.55795 # aprx P(B)
[1] 0.48821 # aprx answer P(A & B)
Because of the order in which I have plotted the 100,000 points below, it is the percentage
of yellow ones that you want, as in your diagram. (About 0.488.)
plot(x, y, pch=".")
points(x[B], y[B], pch=".", col="red") # Read [ ] as "such that"
points(x[A], y[A], pch=".", col="green2")
points(x[A&B], y[A&B], pch=".", col="yellow")
For an exact solution it may be easiest to subtract the area of the green
'nibble' from 1/2. (As @copperhat has suggested.)
answered Aug 1 at 2:30
BruceET
33.1k61440
Comment: You have a path to the answer; this is just a way to confirm
you are on the right track without actually working the problem.
I assume you have independent random variables $X$ and $Y,$ each distributed $mathsfUnif(0, 1).$ And you seek $P(X+Y le 1, XY le 2/9),$ where as usual the comma means intersection.
Here is a simulation in R statistical software of 100,000 independent $X$'s and $Y$'s that illustrates the problem and your approach. [This method of getting
the answer by simulation is sometimes called 'Monte Carlo integration'. It is
mainly used for problems that are too messy to handle with ordinary Riemann integration (often in more than 2 dimensions).
You should use ordinary integration for your answer.]
set.seed(731); m = 10^5; x = runif(m); y = runif(m)
A = (x + y <= 1); B = (x*y <= 2/9) # A and B are logical vectors of TRUEs and FALSEs.
mean(A); mean(B); mean(A & B) # mean of a logical vector is its proportion of TRUEs.
[1] 0.50086 # aprx P(A) = 1/2
[1] 0.55795 # aprx P(B)
[1] 0.48821 # aprx answer P(A & B)
Because of the order in which I have plotted the 100,000 points below, it is the percentage
of yellow ones that you want, as in your diagram. (About 0.488.)
plot(x, y, pch=".")
points(x[B], y[B], pch=".", col="red") # Read [ ] as "such that"
points(x[A], y[A], pch=".", col="green2")
points(x[A&B], y[A&B], pch=".", col="yellow")
For an exact solution it may be easiest to subtract the area of the green
'nibble' from 1/2. (As @copperhat has suggested.)
answered Aug 1 at 2:30
BruceET
33.1k61440
edited Aug 1 at 2:53
answered Aug 1 at 2:30
BruceET
33.1k61440
answered Aug 1 at 2:30
BruceET
33.1k61440
answered Aug 1 at 2:30
BruceET
33.1k61440
33.1k61440
add a comment |Â
add a comment |Â
up vote
2
down vote
One way is to compute the area of the triangle ($=1 over 2$) and subtract the area between
the red and blue lines.
The red and blue lines are $y= 2 over 9x$ and $y = 1-x$. Equating and
solving the resulting quadratic gives the bounds of integration
$1 over 2 (1 pm 1 over 3)$. Hence the little area is
$int_1 over 2 (1 - 1 over 3)^1 over 2 (1 + 1 over 3) (1-x-2 over 9x) dx = 1 over 6-2 over 9 ln 2 $.
The probability is then $1 over 3+2 over 9 ln 2 approx 0.487$.
answered Aug 1 at 5:42
copper.hat
122k557155
1
Simulation with a million iterations (too many for a nice figure) agrees with this to three places. (+1) Also clear and to the point.
– BruceET
Aug 2 at 17:16
That's what I calculated. I don't know how I got confused with you in chat. :)
– Damn1o1
Aug 4 at 8:24
add a comment |Â
up vote
2
down vote
One way is to compute the area of the triangle ($=1 over 2$) and subtract the area between
the red and blue lines.
The red and blue lines are $y= 2 over 9x$ and $y = 1-x$. Equating and
solving the resulting quadratic gives the bounds of integration
$1 over 2 (1 pm 1 over 3)$. Hence the little area is
$int_1 over 2 (1 - 1 over 3)^1 over 2 (1 + 1 over 3) (1-x-2 over 9x) dx = 1 over 6-2 over 9 ln 2 $.
The probability is then $1 over 3+2 over 9 ln 2 approx 0.487$.
answered Aug 1 at 5:42
copper.hat
122k557155
1
Simulation with a million iterations (too many for a nice figure) agrees with this to three places. (+1) Also clear and to the point.
– BruceET
Aug 2 at 17:16
That's what I calculated. I don't know how I got confused with you in chat. :)
– Damn1o1
Aug 4 at 8:24
add a comment |Â
up vote
2
down vote
up vote
2
down vote
One way is to compute the area of the triangle ($=1 over 2$) and subtract the area between
the red and blue lines.
The red and blue lines are $y= 2 over 9x$ and $y = 1-x$. Equating and
solving the resulting quadratic gives the bounds of integration
$1 over 2 (1 pm 1 over 3)$. Hence the little area is
$int_1 over 2 (1 - 1 over 3)^1 over 2 (1 + 1 over 3) (1-x-2 over 9x) dx = 1 over 6-2 over 9 ln 2 $.
The probability is then $1 over 3+2 over 9 ln 2 approx 0.487$.
answered Aug 1 at 5:42
copper.hat
122k557155
One way is to compute the area of the triangle ($=1 over 2$) and subtract the area between
the red and blue lines.
The red and blue lines are $y= 2 over 9x$ and $y = 1-x$. Equating and
solving the resulting quadratic gives the bounds of integration
$1 over 2 (1 pm 1 over 3)$. Hence the little area is
$int_1 over 2 (1 - 1 over 3)^1 over 2 (1 + 1 over 3) (1-x-2 over 9x) dx = 1 over 6-2 over 9 ln 2 $.
The probability is then $1 over 3+2 over 9 ln 2 approx 0.487$.
answered Aug 1 at 5:42
copper.hat
122k557155
answered Aug 1 at 5:42
copper.hat
122k557155
answered Aug 1 at 5:42
copper.hat
122k557155
answered Aug 1 at 5:42
copper.hat
122k557155
122k557155
1
Simulation with a million iterations (too many for a nice figure) agrees with this to three places. (+1) Also clear and to the point.
– BruceET
Aug 2 at 17:16
That's what I calculated. I don't know how I got confused with you in chat. :)
– Damn1o1
Aug 4 at 8:24
add a comment |Â
1
Simulation with a million iterations (too many for a nice figure) agrees with this to three places. (+1) Also clear and to the point.
– BruceET
Aug 2 at 17:16
That's what I calculated. I don't know how I got confused with you in chat. :)
– Damn1o1
Aug 4 at 8:24
1
1
Simulation with a million iterations (too many for a nice figure) agrees with this to three places. (+1) Also clear and to the point.
– BruceET
Aug 2 at 17:16
Simulation with a million iterations (too many for a nice figure) agrees with this to three places. (+1) Also clear and to the point.
– BruceET
Aug 2 at 17:16
That's what I calculated. I don't know how I got confused with you in chat. :)
– Damn1o1
Aug 4 at 8:24
That's what I calculated. I don't know how I got confused with you in chat. :)
– Damn1o1
Aug 4 at 8:24
add a comment |Â
up vote
1
down vote
Be $X,Y sim U(0,1)$ assuming $X perp Y$ and let $A = X + Y$ and $B=XY$. We’re trying to evaluate:
$$mathrmPleft[A leq 1 cap B leq frac29right]$$
As other answers have already pointed out it is handy to plot the sets. However even if it is correct to say “the probability I’m looking for is the value of that area†given this scenario, I’d like to take a deeper look for why it is that.
Be $D subset mathbbR^2$ the set we’re interested into, which is defined as:
$$D = left(x,y):x+yleq1,xyleqfrac29,xgeq0,ygeq0right$$
Then:
$$
mathrmPleft[A leq 1 cap B leq frac29right]=iint_DmathrmPleft[X = x cap Y = yright],dx,dy
$$
We’re considering every point $(x_0,y_0)$ that satisfies our conditions, that is when $X=x_0$ and $Y=y_0$ are the coordinates of a point inside $D$ at the same time.
Since $Xperp Y$ then:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= 1 cdot 1 \
&= 1
endalign
$$
Now we see why $mathrmPleft[A leq 1 cap B leq frac29right]$ is that area, because:
$$
beginalign
mathrmPleft[A leq 1 cap B leq frac29right]=iint_Ddx,dy=textArea,(D)
endalign
$$
Extra work - while other answers have already found the number you need I’d like to verify what I wrote before. Let’s say that $Xsim U(0,2)$ and $Ysim U(1,4)$. We want to evaluate:
$$
beginalign
mathrmPleft[X+Y leq frac72 cap 2X+Y geq 3right]
endalign
$$
Time to define $D$:
$$D = left(x,y):x+yleqfrac72,2x+ygeq3,0leq x<2,1leq y<4right$$
We’ll plot it using IPython and Matplotlib.
import numpy as np
import matplotlib as mpl
N = 10**6
X = np.random.uniform(0, 2, N)
Y = np.random.uniform(1, 4, N)
A = (X + Y <= 4)
B = (2*X + Y >= 3)
np.mean(A&B) # ~= 0.333
pp.plot(X[A&B], Y[A&B], '.'); pp.show()
We’re going to reference this later to set up the integral. We need to compute now:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= frac12cdotfrac13 \
&= frac16
endalign
$$
Let’s begin:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16iint_Ddx,dy
$$
Looking at $D$ we can easily see a trapezoid from which a triangle has been dropped, then:
$$
iint_Ddx,dy = 2fracfrac12+frac522-frac1cdot 22 = 2
$$
Finally:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16cdot 2 = frac13 simeq 0.333
$$
answered Aug 2 at 1:30
Giulio Scattolin
15619
You should recheck your answer. Other have given the answer so you can confirm it.
– Damn1o1
Aug 4 at 8:25
@Damn1o1 I apologize, you want me to provide the answer using the method I presented?
– Giulio Scattolin
Aug 4 at 9:51
Yes, that will be fine. I want to learn things. A different approach will be better.
– Damn1o1
Aug 4 at 9:53
@Damn1o1 Actually copper.hat's computations are what I would do now, that's way I stopped there. At the moment I don't believe there is another way to get to the answer to your problem without evaluating that area somehow: that area (in this problem) is the probability you're interested to, so any "different approach" will bring you to evaluate that area differently. Instead, you could obtain a different formula starting from the $iint$ but I don't see any benefit since that would be more a "geometric" problem, let's say, more than a "statistic" one.
– Giulio Scattolin
Aug 4 at 10:04
add a comment |Â
up vote
1
down vote
Be $X,Y sim U(0,1)$ assuming $X perp Y$ and let $A = X + Y$ and $B=XY$. We’re trying to evaluate:
$$mathrmPleft[A leq 1 cap B leq frac29right]$$
As other answers have already pointed out it is handy to plot the sets. However even if it is correct to say “the probability I’m looking for is the value of that area†given this scenario, I’d like to take a deeper look for why it is that.
Be $D subset mathbbR^2$ the set we’re interested into, which is defined as:
$$D = left(x,y):x+yleq1,xyleqfrac29,xgeq0,ygeq0right$$
Then:
$$
mathrmPleft[A leq 1 cap B leq frac29right]=iint_DmathrmPleft[X = x cap Y = yright],dx,dy
$$
We’re considering every point $(x_0,y_0)$ that satisfies our conditions, that is when $X=x_0$ and $Y=y_0$ are the coordinates of a point inside $D$ at the same time.
Since $Xperp Y$ then:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= 1 cdot 1 \
&= 1
endalign
$$
Now we see why $mathrmPleft[A leq 1 cap B leq frac29right]$ is that area, because:
$$
beginalign
mathrmPleft[A leq 1 cap B leq frac29right]=iint_Ddx,dy=textArea,(D)
endalign
$$
Extra work - while other answers have already found the number you need I’d like to verify what I wrote before. Let’s say that $Xsim U(0,2)$ and $Ysim U(1,4)$. We want to evaluate:
$$
beginalign
mathrmPleft[X+Y leq frac72 cap 2X+Y geq 3right]
endalign
$$
Time to define $D$:
$$D = left(x,y):x+yleqfrac72,2x+ygeq3,0leq x<2,1leq y<4right$$
We’ll plot it using IPython and Matplotlib.
import numpy as np
import matplotlib as mpl
N = 10**6
X = np.random.uniform(0, 2, N)
Y = np.random.uniform(1, 4, N)
A = (X + Y <= 4)
B = (2*X + Y >= 3)
np.mean(A&B) # ~= 0.333
pp.plot(X[A&B], Y[A&B], '.'); pp.show()
We’re going to reference this later to set up the integral. We need to compute now:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= frac12cdotfrac13 \
&= frac16
endalign
$$
Let’s begin:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16iint_Ddx,dy
$$
Looking at $D$ we can easily see a trapezoid from which a triangle has been dropped, then:
$$
iint_Ddx,dy = 2fracfrac12+frac522-frac1cdot 22 = 2
$$
Finally:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16cdot 2 = frac13 simeq 0.333
$$
answered Aug 2 at 1:30
Giulio Scattolin
15619
You should recheck your answer. Other have given the answer so you can confirm it.
– Damn1o1
Aug 4 at 8:25
@Damn1o1 I apologize, you want me to provide the answer using the method I presented?
– Giulio Scattolin
Aug 4 at 9:51
Yes, that will be fine. I want to learn things. A different approach will be better.
– Damn1o1
Aug 4 at 9:53
@Damn1o1 Actually copper.hat's computations are what I would do now, that's way I stopped there. At the moment I don't believe there is another way to get to the answer to your problem without evaluating that area somehow: that area (in this problem) is the probability you're interested to, so any "different approach" will bring you to evaluate that area differently. Instead, you could obtain a different formula starting from the $iint$ but I don't see any benefit since that would be more a "geometric" problem, let's say, more than a "statistic" one.
– Giulio Scattolin
Aug 4 at 10:04
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Be $X,Y sim U(0,1)$ assuming $X perp Y$ and let $A = X + Y$ and $B=XY$. We’re trying to evaluate:
$$mathrmPleft[A leq 1 cap B leq frac29right]$$
As other answers have already pointed out it is handy to plot the sets. However even if it is correct to say “the probability I’m looking for is the value of that area†given this scenario, I’d like to take a deeper look for why it is that.
Be $D subset mathbbR^2$ the set we’re interested into, which is defined as:
$$D = left(x,y):x+yleq1,xyleqfrac29,xgeq0,ygeq0right$$
Then:
$$
mathrmPleft[A leq 1 cap B leq frac29right]=iint_DmathrmPleft[X = x cap Y = yright],dx,dy
$$
We’re considering every point $(x_0,y_0)$ that satisfies our conditions, that is when $X=x_0$ and $Y=y_0$ are the coordinates of a point inside $D$ at the same time.
Since $Xperp Y$ then:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= 1 cdot 1 \
&= 1
endalign
$$
Now we see why $mathrmPleft[A leq 1 cap B leq frac29right]$ is that area, because:
$$
beginalign
mathrmPleft[A leq 1 cap B leq frac29right]=iint_Ddx,dy=textArea,(D)
endalign
$$
Extra work - while other answers have already found the number you need I’d like to verify what I wrote before. Let’s say that $Xsim U(0,2)$ and $Ysim U(1,4)$. We want to evaluate:
$$
beginalign
mathrmPleft[X+Y leq frac72 cap 2X+Y geq 3right]
endalign
$$
Time to define $D$:
$$D = left(x,y):x+yleqfrac72,2x+ygeq3,0leq x<2,1leq y<4right$$
We’ll plot it using IPython and Matplotlib.
import numpy as np
import matplotlib as mpl
N = 10**6
X = np.random.uniform(0, 2, N)
Y = np.random.uniform(1, 4, N)
A = (X + Y <= 4)
B = (2*X + Y >= 3)
np.mean(A&B) # ~= 0.333
pp.plot(X[A&B], Y[A&B], '.'); pp.show()
We’re going to reference this later to set up the integral. We need to compute now:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= frac12cdotfrac13 \
&= frac16
endalign
$$
Let’s begin:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16iint_Ddx,dy
$$
Looking at $D$ we can easily see a trapezoid from which a triangle has been dropped, then:
$$
iint_Ddx,dy = 2fracfrac12+frac522-frac1cdot 22 = 2
$$
Finally:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16cdot 2 = frac13 simeq 0.333
$$
answered Aug 2 at 1:30
Giulio Scattolin
15619
Be $X,Y sim U(0,1)$ assuming $X perp Y$ and let $A = X + Y$ and $B=XY$. We’re trying to evaluate:
$$mathrmPleft[A leq 1 cap B leq frac29right]$$
As other answers have already pointed out it is handy to plot the sets. However even if it is correct to say “the probability I’m looking for is the value of that area†given this scenario, I’d like to take a deeper look for why it is that.
Be $D subset mathbbR^2$ the set we’re interested into, which is defined as:
$$D = left(x,y):x+yleq1,xyleqfrac29,xgeq0,ygeq0right$$
Then:
$$
mathrmPleft[A leq 1 cap B leq frac29right]=iint_DmathrmPleft[X = x cap Y = yright],dx,dy
$$
We’re considering every point $(x_0,y_0)$ that satisfies our conditions, that is when $X=x_0$ and $Y=y_0$ are the coordinates of a point inside $D$ at the same time.
Since $Xperp Y$ then:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= 1 cdot 1 \
&= 1
endalign
$$
Now we see why $mathrmPleft[A leq 1 cap B leq frac29right]$ is that area, because:
$$
beginalign
mathrmPleft[A leq 1 cap B leq frac29right]=iint_Ddx,dy=textArea,(D)
endalign
$$
Extra work - while other answers have already found the number you need I’d like to verify what I wrote before. Let’s say that $Xsim U(0,2)$ and $Ysim U(1,4)$. We want to evaluate:
$$
beginalign
mathrmPleft[X+Y leq frac72 cap 2X+Y geq 3right]
endalign
$$
Time to define $D$:
$$D = left(x,y):x+yleqfrac72,2x+ygeq3,0leq x<2,1leq y<4right$$
We’ll plot it using IPython and Matplotlib.
import numpy as np
import matplotlib as mpl
N = 10**6
X = np.random.uniform(0, 2, N)
Y = np.random.uniform(1, 4, N)
A = (X + Y <= 4)
B = (2*X + Y >= 3)
np.mean(A&B) # ~= 0.333
pp.plot(X[A&B], Y[A&B], '.'); pp.show()
We’re going to reference this later to set up the integral. We need to compute now:
$$
beginalign
mathrmPleft[X = x cap Y = yright] &= mathrmPleft[X = xright],mathrmPleft[Y = yright] \
&= frac12cdotfrac13 \
&= frac16
endalign
$$
Let’s begin:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16iint_Ddx,dy
$$
Looking at $D$ we can easily see a trapezoid from which a triangle has been dropped, then:
$$
iint_Ddx,dy = 2fracfrac12+frac522-frac1cdot 22 = 2
$$
Finally:
$$
iint_DmathrmPleft[X = x cap Y = yright]dx,dy = frac16cdot 2 = frac13 simeq 0.333
$$
answered Aug 2 at 1:30
Giulio Scattolin
15619
edited Aug 4 at 10:21
answered Aug 2 at 1:30
Giulio Scattolin
15619
answered Aug 2 at 1:30
Giulio Scattolin
15619
answered Aug 2 at 1:30
Giulio Scattolin
15619
15619
You should recheck your answer. Other have given the answer so you can confirm it.
– Damn1o1
Aug 4 at 8:25
@Damn1o1 I apologize, you want me to provide the answer using the method I presented?
– Giulio Scattolin
Aug 4 at 9:51
Yes, that will be fine. I want to learn things. A different approach will be better.
– Damn1o1
Aug 4 at 9:53
@Damn1o1 Actually copper.hat's computations are what I would do now, that's way I stopped there. At the moment I don't believe there is another way to get to the answer to your problem without evaluating that area somehow: that area (in this problem) is the probability you're interested to, so any "different approach" will bring you to evaluate that area differently. Instead, you could obtain a different formula starting from the $iint$ but I don't see any benefit since that would be more a "geometric" problem, let's say, more than a "statistic" one.
– Giulio Scattolin
Aug 4 at 10:04
add a comment |Â
You should recheck your answer. Other have given the answer so you can confirm it.
– Damn1o1
Aug 4 at 8:25
@Damn1o1 I apologize, you want me to provide the answer using the method I presented?
– Giulio Scattolin
Aug 4 at 9:51
Yes, that will be fine. I want to learn things. A different approach will be better.
– Damn1o1
Aug 4 at 9:53
@Damn1o1 Actually copper.hat's computations are what I would do now, that's way I stopped there. At the moment I don't believe there is another way to get to the answer to your problem without evaluating that area somehow: that area (in this problem) is the probability you're interested to, so any "different approach" will bring you to evaluate that area differently. Instead, you could obtain a different formula starting from the $iint$ but I don't see any benefit since that would be more a "geometric" problem, let's say, more than a "statistic" one.
– Giulio Scattolin
Aug 4 at 10:04
You should recheck your answer. Other have given the answer so you can confirm it.
– Damn1o1
Aug 4 at 8:25
You should recheck your answer. Other have given the answer so you can confirm it.
– Damn1o1
Aug 4 at 8:25
@Damn1o1 I apologize, you want me to provide the answer using the method I presented?
– Giulio Scattolin
Aug 4 at 9:51
@Damn1o1 I apologize, you want me to provide the answer using the method I presented?
– Giulio Scattolin
Aug 4 at 9:51
Yes, that will be fine. I want to learn things. A different approach will be better.
– Damn1o1
Aug 4 at 9:53
Yes, that will be fine. I want to learn things. A different approach will be better.
– Damn1o1
Aug 4 at 9:53
@Damn1o1 Actually copper.hat's computations are what I would do now, that's way I stopped there. At the moment I don't believe there is another way to get to the answer to your problem without evaluating that area somehow: that area (in this problem) is the probability you're interested to, so any "different approach" will bring you to evaluate that area differently. Instead, you could obtain a different formula starting from the $iint$ but I don't see any benefit since that would be more a "geometric" problem, let's say, more than a "statistic" one.
– Giulio Scattolin
Aug 4 at 10:04
@Damn1o1 Actually copper.hat's computations are what I would do now, that's way I stopped there. At the moment I don't believe there is another way to get to the answer to your problem without evaluating that area somehow: that area (in this problem) is the probability you're interested to, so any "different approach" will bring you to evaluate that area differently. Instead, you could obtain a different formula starting from the $iint$ but I don't see any benefit since that would be more a "geometric" problem, let's say, more than a "statistic" one.
– Giulio Scattolin
Aug 4 at 10:04
add a comment |Â
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1
You need to specify the distribution of the two randomly chosen numbers. Are they uniformly distributed?
– Theoretical Economist
Jul 31 at 15:36
Presumably your are using a uniform distribution in which case your approach is a good one.
– copper.hat
Jul 31 at 15:37
@TheoreticalEconomist I assumed that here.
– Damn1o1
Jul 31 at 15:37
1
There is a triangle and a little half banana like area. The probability is the area of the triangle less the half banana area.
– copper.hat
Jul 31 at 15:55
1
I'm confused by what you are asking. You want to compute the yellow area.
– copper.hat
Jul 31 at 15:57