Is symmetry necessary for solving triple integrals in probability calculations?

[jaumzaum]

Homework Statement

Consider T to be the set of triangles from R3+.
i.e. T = {(a, b, c) | max(a, b, c)<(a+b+c)/2}
Consider an approximately equilateral triangle to be a triangle in which the major side does not exceed the minor side in 10%. Consider E to be the set of approximately equilateral triangles.
i.e. E = {(a, b, c) | max(a, b, c)<1.1*min(a, b, c)}
What is the chance that a given triangle (i.e. picked from T) is approximately equilateral (i.e. belongs to E)

Relevant Equations

I don't think there are any specific equation to use.

I just started learning triple integrals. I don't know if this is right (I'm only concerned about the limits of the Integral)
Consider the case $a>=b>=c$

$$ P = \lim_{M \rightarrow +\infty} \frac { \int_{a=0}^M \int_{b=\frac a {1.1}}^a \int_{c= \frac a {1.1}}^b \, da \, db \, dc} { \int_{a=0}^M \int_{b=0}^a \int_{c=0}^b \, da \, db \, dc} = \frac 1 {121} = 0.00826 $$

 

[PeroK]

Homework Statement: Consider T to be the set of triangles from R3+.
i.e. T = {(a, b, c) | max(a, b, c)<(a+b+c)/2}
Consider an approximately equilateral triangle to be a triangle in which the major side does not exceed the minor side in 10%. Consider E to be the set of approximately equilateral triangles.
i.e. E = {(a, b, c) | max(a, b, c)<1.1*min(a, b, c)}
What is the chance that a given triangle (i.e. picked from T) is approximately equilateral (i.e. belongs to E)
Homework Equations: I don't think there are any specific equation to use.

I just started learning triple integrals. I don't know if this is right (I'm only concerned about the limits of the Integral)
Consider the case $a>=b>=c$

$$ P = \lim_{M \rightarrow +\infty} \frac { \int_{a=0}^M \int_{b=\frac a {1.1}}^a \int_{c= \frac a {1.1}}^b \, da \, db \, dc} { \int_{a=0}^M \int_{b=0}^a \int_{c=0}^b \, da \, db \, dc} = \frac 1 {121} = 0.00826 $$

How do you ensure that $a > b > c$? One way is to choose the three sides uniformly at random on some range ($(0, M)$ in your notation) and simply reject the choice if they don't come in descending order. Then you can argue that the probability they form a triangle and the probability they form a nearly equilateral triangle are the same for your restricted sample space as for the full sample space, including all the ones you rejected. So far, so good.

But, now $a, b, c$ are no longer distibuted uniformly on the given intervals. $a$ is more likely to be large ( close to $M$) than small (close to $0$).

So, I don't think you can have the constraint that $a > b >c$ and retain the uniform distribution.

Second point (again assuming $a > b > c$): for the numbers to form a triangle, you have $c > a-b$. The limits on the integral in the denominator for $c$, therefore, are not correct.

A very tricky problem!

 

[jaumzaum]

How do you ensure that $a > b > c$? One way is to choose the three sides uniformly at random on some range ($(0, M)$ in your notation) and simply reject the choice if they don't come in descending order. Then you can argue that the probability they form a triangle and the probability they form a nearly equilateral triangle are the same for your restricted sample space as for the full sample space, including all the ones you rejected. So far, so good.

But, now $a, b, c$ are no longer distibuted uniformly on the given intervals. $a$ is more likely to be large ( close to $M$) than small (close to $0$).

So, I don't think you can have the constraint that $a > b >c$ and retain the uniform distribution.

I thought about this when I first tried to answer the problem.
Let's not consider that a>=b>=c, but any triple (a, b, c) instead.
As we have 6 different permutations of (a, b, c), and independently of a, b or c being the major number the numerator would be the same. So if we assume (a, b, c) belongs to (0, M)^3, the numerator would be 6 times what we first calculated. The denominator would also be 6 times, and so the fraction would be the same.

The problem with that (and it was one of my doubts) is that we considered (a, b, c) belonging to (0, M)^3 (a cube), with M approaching infinity. But we could also consider (a, b, c) belonging to (0, M) x (0, N) x (0, O) (a parallelepiped).
That way we would get a non-solvable limit.

Second point (again assuming $a > b > c$): for the numbers to form a triangle, you have $c > a-b$. The limits on the integral in the denominator for $c$, therefore, are not correct.

You are right! I will change it.

 

[jaumzaum]

$$ P = \lim_{M \rightarrow +\infty} \frac { \int_{a=0}^M \int_{b=\frac a {1.1}}^a \int_{c= \frac a {1.1}}^b \, da \, db \, dc} { \int_{a=0}^M \int_{b=a/2}^a \int_{c=a-b}^b \, da \, db \, dc} = \frac 2 {121} = 0.01653 $$

 

[PeroK]

The problem with that (and it was one of my doubts) is that we considered (a, b, c) belonging to (0, M)^3 (a cube), with M approaching infinity. But we could also consider (a, b, c) belonging to (0, M) x (0, N) x (0, O) (a parallelepiped).
That way we would get a non-solvable limit.

These problems are always going to depend on how you choose the numbers. You could use spherical coordinates, for example, and get an entirely different answer.

Logically you require symmetry of $a, b, c$, as you have done.