# Let $x, y,$ and $z$ be selected uniformly and independently at random over the interval $(0, 3)$. What is the probability that $x + y + z > 1$? [closed]

I am having trouble finding the number of possible combinations for this problem:

Question:

Let $$x$$, $$y$$, and $$z$$ be selected uniformly and independently at random over the interval $$(0, 3)$$. What is the probability that $$x + y + z > 1?$$

## closed as off-topic by NCh, Shailesh, mrtaurho, Song, Parcly TaxelFeb 28 at 8:23

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• What have you tried? – Parcly Taxel Feb 27 at 13:05
• Are $x, y, z$ are integers? – Abhinav Feb 27 at 13:46

You select a uniformly distributed random point $$(x,y,z)$$ in the cube $$C:=[0,3]^3$$. Draw a figure! Find the set $$B\subset C$$ of "bad" points (where $$x+y+z\leq 1$$), compute the probability $$P(B)$$, and finally $$P(C\setminus B)$$.

An analytic approach

$$1-\frac {\int_0^1\int_0^{1-z}\int_0^{1-y-z} \ dx \ dy\ dz}{\int_0^3\int_0^3\int_0^3 \ dx \ dy\ dz}$$

Or if you prefer

$$\frac {\int_0^3\int_{1-z}^{3}\int_{1-y-z}^{3} \ dx \ dy\ dz}{\int_0^3\int_0^3\int_0^3 \ dx \ dy\ dz}$$

A Geometric approach

$$x+y+z \le 1$$ with $$x,y,z > 0$$ defines a tetrahedron -- the volume under the plane in the first quadrant.

What is the volume of the cube less the tetrahedron compared to the volume of the cube?