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


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 Taxel Feb 28 at 8:23

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  • 3
    $\begingroup$ What have you tried? $\endgroup$ – Parcly Taxel Feb 27 at 13:05
  • $\begingroup$ Are $x, y, z$ are integers? $\endgroup$ – 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?


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