Mixing
A problem on conditional geometric probability
Clash Royale CLAN TAG#URR8PPP
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The point (x, y) is chosen randomly in the unit square. What is the conditional probability of $x^2+y^2 leq 1/4$ given that $xy leq 1/16$
I started solving this and while calculating I got some very unpleasant numbers but the problem is not marked as difficult in this book where I found it. So I start suspecting that I am having some conceptual mistake.
The two curves intersect at $x=sqrt2+sqrt 3 /4$ and at $x=sqrt2-sqrt 3 /4$
Now I have to find a few areas via definite integrals. Is that what this problem is about?!
The integral of $sqrt1/4-x^2 $ is very unpleasant, it seems. So?!
Is there some smarter / nicer approach?
probability probability-theory
edited Jul 17 at 6:48
Martin Roberts
1,189318
asked Jul 16 at 23:23
peter.petrov
5,321721
add a comment |Â
up vote
0
down vote
favorite
The point (x, y) is chosen randomly in the unit square. What is the conditional probability of $x^2+y^2 leq 1/4$ given that $xy leq 1/16$
I started solving this and while calculating I got some very unpleasant numbers but the problem is not marked as difficult in this book where I found it. So I start suspecting that I am having some conceptual mistake.
The two curves intersect at $x=sqrt2+sqrt 3 /4$ and at $x=sqrt2-sqrt 3 /4$
Now I have to find a few areas via definite integrals. Is that what this problem is about?!
The integral of $sqrt1/4-x^2 $ is very unpleasant, it seems. So?!
Is there some smarter / nicer approach?
probability probability-theory
edited Jul 17 at 6:48
Martin Roberts
1,189318
asked Jul 16 at 23:23
peter.petrov
5,321721
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The point (x, y) is chosen randomly in the unit square. What is the conditional probability of $x^2+y^2 leq 1/4$ given that $xy leq 1/16$
I started solving this and while calculating I got some very unpleasant numbers but the problem is not marked as difficult in this book where I found it. So I start suspecting that I am having some conceptual mistake.
The two curves intersect at $x=sqrt2+sqrt 3 /4$ and at $x=sqrt2-sqrt 3 /4$
Now I have to find a few areas via definite integrals. Is that what this problem is about?!
The integral of $sqrt1/4-x^2 $ is very unpleasant, it seems. So?!
Is there some smarter / nicer approach?
probability probability-theory
edited Jul 17 at 6:48
Martin Roberts
1,189318
asked Jul 16 at 23:23
peter.petrov
5,321721
The point (x, y) is chosen randomly in the unit square. What is the conditional probability of $x^2+y^2 leq 1/4$ given that $xy leq 1/16$
I started solving this and while calculating I got some very unpleasant numbers but the problem is not marked as difficult in this book where I found it. So I start suspecting that I am having some conceptual mistake.
The two curves intersect at $x=sqrt2+sqrt 3 /4$ and at $x=sqrt2-sqrt 3 /4$
Now I have to find a few areas via definite integrals. Is that what this problem is about?!
The integral of $sqrt1/4-x^2 $ is very unpleasant, it seems. So?!
Is there some smarter / nicer approach?
probability probability-theory
edited Jul 17 at 6:48
Martin Roberts
1,189318
asked Jul 16 at 23:23
peter.petrov
5,321721
edited Jul 17 at 6:48
Martin Roberts
1,189318
edited Jul 17 at 6:48
Martin Roberts
1,189318
edited Jul 17 at 6:48
Martin Roberts
1,189318
1,189318
asked Jul 16 at 23:23
peter.petrov
5,321721
asked Jul 16 at 23:23
peter.petrov
5,321721
asked Jul 16 at 23:23
peter.petrov
5,321721
5,321721
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
$xy leq frac116$ iff $2xy leq frac18$.
Conditional on this,
$x^2 + y^2 leq frac14$ iff $x^2 + 2xy + y^2 = (x+y)^2 leq frac38$ iff $x+y leq sqrtfrac38$.
Now do you know how to calculate the probability that the sum of two random numbers is less than a given number?
answered Jul 16 at 23:43
Badam Baplan
3,331721
Good, however one of the original inequalities shall be kept, of course. So it is the area of the circle remaining below that line.
– G Cab
Jul 17 at 0:01
add a comment |Â
up vote
-1
down vote
Suppose that $xyle frac116.$ Then $x^2+y^2le frac14iff (x+y)^2le frac38$ This should lead to an easier way to approach it.
answered Jul 16 at 23:44
saulspatz
10.7k21323
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$xy leq frac116$ iff $2xy leq frac18$.
Conditional on this,
$x^2 + y^2 leq frac14$ iff $x^2 + 2xy + y^2 = (x+y)^2 leq frac38$ iff $x+y leq sqrtfrac38$.
Now do you know how to calculate the probability that the sum of two random numbers is less than a given number?
answered Jul 16 at 23:43
Badam Baplan
3,331721
Good, however one of the original inequalities shall be kept, of course. So it is the area of the circle remaining below that line.
– G Cab
Jul 17 at 0:01
add a comment |Â
up vote
2
down vote
$xy leq frac116$ iff $2xy leq frac18$.
Conditional on this,
$x^2 + y^2 leq frac14$ iff $x^2 + 2xy + y^2 = (x+y)^2 leq frac38$ iff $x+y leq sqrtfrac38$.
Now do you know how to calculate the probability that the sum of two random numbers is less than a given number?
answered Jul 16 at 23:43
Badam Baplan
3,331721
Good, however one of the original inequalities shall be kept, of course. So it is the area of the circle remaining below that line.
– G Cab
Jul 17 at 0:01
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$xy leq frac116$ iff $2xy leq frac18$.
Conditional on this,
$x^2 + y^2 leq frac14$ iff $x^2 + 2xy + y^2 = (x+y)^2 leq frac38$ iff $x+y leq sqrtfrac38$.
Now do you know how to calculate the probability that the sum of two random numbers is less than a given number?
answered Jul 16 at 23:43
Badam Baplan
3,331721
$xy leq frac116$ iff $2xy leq frac18$.
Conditional on this,
$x^2 + y^2 leq frac14$ iff $x^2 + 2xy + y^2 = (x+y)^2 leq frac38$ iff $x+y leq sqrtfrac38$.
Now do you know how to calculate the probability that the sum of two random numbers is less than a given number?
answered Jul 16 at 23:43
Badam Baplan
3,331721
answered Jul 16 at 23:43
Badam Baplan
3,331721
answered Jul 16 at 23:43
Badam Baplan
3,331721
answered Jul 16 at 23:43
Badam Baplan
3,331721
3,331721
Good, however one of the original inequalities shall be kept, of course. So it is the area of the circle remaining below that line.
– G Cab
Jul 17 at 0:01
add a comment |Â
Good, however one of the original inequalities shall be kept, of course. So it is the area of the circle remaining below that line.
– G Cab
Jul 17 at 0:01
Good, however one of the original inequalities shall be kept, of course. So it is the area of the circle remaining below that line.
– G Cab
Jul 17 at 0:01
Good, however one of the original inequalities shall be kept, of course. So it is the area of the circle remaining below that line.
– G Cab
Jul 17 at 0:01
add a comment |Â
up vote
-1
down vote
Suppose that $xyle frac116.$ Then $x^2+y^2le frac14iff (x+y)^2le frac38$ This should lead to an easier way to approach it.
answered Jul 16 at 23:44
saulspatz
10.7k21323
add a comment |Â
up vote
-1
down vote
Suppose that $xyle frac116.$ Then $x^2+y^2le frac14iff (x+y)^2le frac38$ This should lead to an easier way to approach it.
answered Jul 16 at 23:44
saulspatz
10.7k21323
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
Suppose that $xyle frac116.$ Then $x^2+y^2le frac14iff (x+y)^2le frac38$ This should lead to an easier way to approach it.
answered Jul 16 at 23:44
saulspatz
10.7k21323
Suppose that $xyle frac116.$ Then $x^2+y^2le frac14iff (x+y)^2le frac38$ This should lead to an easier way to approach it.
answered Jul 16 at 23:44
saulspatz
10.7k21323
answered Jul 16 at 23:44
saulspatz
10.7k21323
answered Jul 16 at 23:44
saulspatz
10.7k21323
answered Jul 16 at 23:44
saulspatz
10.7k21323
10.7k21323
add a comment |Â
add a comment |Â
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