Lost in a cube

Geometry Level 4

A point is chosen uniformly at random inside the cube defined by:

$0 < x < a$
$0 < y < a$
$0 < z < a$

What is the probability that $x+y+z < a$ ?

Give your answer to 3 decimal places.

The answer is 0.167.

1 solution

Brian Charlesworth
Apr 2, 2017

The desired probability will be $p = \dfrac{A}{a^{3}}$ , where $A$ is the volume of the region $R$ bounded by the $xy, xz$ and $yz$ planes and the plane $x + y + z = a$ , and $a^{3}$ is the volume of the cube.

Now $R$ is a tetrahedron with a triangular base of area $\dfrac{a^{2}}{2}$ and height $a$ , so

$A = \dfrac{1}{3} \times \dfrac{a^{2}}{2} \times a = \dfrac{a^{3}}{6}$ , and thus $p = \dfrac{1}{6} = \boxed{0.167}$ to 3 decimal places.

Follow-up question: True or false: In general, in $\mathbb{R}^{n}$ , if $0 \lt x_{k} \lt a, 1 \le k \le n$ , then the probability that $\displaystyle\sum_{k=1}^{n} x_{k} \lt a$ is $\dfrac{1}{n!}$ .

@Brian Charlesworth Showing that it is $\dfrac{1}{n!}$ ...

I think that can be done with an argument as follows...

For example for six variables, $u, v, w, x, y$ and $z$ we would have something like:

$P = \int_0^1 \int_0^u \int_0^v \int_0^w \int_0^x \int_0^y dz dy dx dw dv du$

$P = \int_0^1 \int_0^u \int_0^v \int_0^w \int_0^x y dy dx dw dv du$

$P = \int_0^1 \int_0^u \int_0^v \int_0^w \dfrac{x^2}{2} dx dw dv du$

$P = \int_0^1 \int_0^u \int_0^v \dfrac{w^3}{6} dw dv du$

$P = \int_0^1 \int_0^u \dfrac{v^4}{24} dv du$

$P = \int_0^1 \dfrac{u^5}{120} du$

$P = \dfrac{1}{720}$

$P = \dfrac{1}{6!}$

Geoff Pilling - 4 years, 2 months ago

Yes, that does it nicely. :)

Brian Charlesworth - 4 years, 2 months ago

@Brian Charlesworth Although this works very nicely for, say, six variables in the interval [0,1] and finding the probability that they add up to one... It seems to be a little more tricky to find the probability that they add up to, say, two... :-/

i.e. My gut instinct was to replace the 1 in the above integral with a 2, but this doesn't quite work...

Geoff Pilling - 4 years, 2 months ago

@Geoff Pilling – No, I guess it doesn't. I find that for three variables, $0 \lt x,y,z \lt 1$ , the probability that $x + y + z \lt 2$ is $3/4$ , while the "integral with a 2" would yield a result in excess of 1.

The probability that $x + y + z \lt 3/2$ is $1/2$ , which makes sense from a symmetry perspective, while for $x + y + z \lt 1/2$ it would be $1/48$ . Some potentially good follow-up questions here ....

Brian Charlesworth - 4 years, 2 months ago

@Brian Charlesworth – I'm having a tough time setting up the integrals for, say 6 variables in [0,1], and finding what is the probability their sum is less than 2.

I'm thinking along the lines of integrating from 0 to 2 for the first variable, u. Then for the second variable, v, we could integrate from max(0,2-u-1) to 2-u. And for the third integrating from max(0,2-u-v-1) to 2-u-v.... etc.

Do you suppose that would work? Perhaps I'll try something along these lines when I get a chance...

I wonder if there is a more intuitive approach?

Geoff Pilling - 4 years, 2 months ago

@Geoff Pilling – Argh. This is difficult. My intuition only extends to 3 variables. More than that and I feel like I'm in a Matrix movie. :p I'll try again later....

Brian Charlesworth - 4 years, 2 months ago

@Brian Charlesworth – Haha... Yup, I haven't got very far with that one either... Interesting how much the whole thing kind of explodes once you want to consider the total to be greater than the possible range of each variable... I'd really like to find a nice elegant approach... But so far... Nothin! :-/

Geoff Pilling - 4 years, 2 months ago

Interesting follow up... I'll need to think about that one!

Geoff Pilling - 4 years, 2 months ago

@Brian Charlesworth Please check. Height is $\frac {a}{\sqrt {3}}$ and the corresponding triangular base area to get the desired result.

Rajen Kapur - 4 years, 2 months ago

I took the base as a triangle in the $xy$ plane and the height as the side along the $z$ -axis to obtain my values.

Brian Charlesworth - 4 years, 2 months ago

Lost in a cube

The answer is 0.167.

1 solution

0 pending reports