If $x$, $y$, and $z$ are random numbers from $0$ to $1$, what is the probability that $x+y+z<1$?
I know that the answer is $\frac{1}{6}$ but I am not sure how to get there. If this problem had two variables, I could graph $x+y<1$ and look at the area to find the answer.
On
Just for the sake of variety:
A geometric answer can be obtained by recognizing that in the unit cube (which has volume $1$), the desired subset is that pyramid whose apex is one of the vertices, and whose base is spanned by the three adjacent vertices.
Since the cube can be decomposed into such four such pyramids (the one described above, plus three more, each with an apex at one of the vertices at distance $\sqrt{2}$ from the first apex) plus a central tetrahedron with twice the volume, each pyramid must have volume $\frac16$. That's the desired probability.
Not necessarily the way you want to go about figuring this out, but it's a sanity check. (In a manner of speaking.)
The following must be satisfied:
$$z \lt1-x-y$$
$$y \lt 1-x$$
$$x \lt 1$$
Thus, we have the integral
$$P(x+y+z \lt 1) = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} 1 dz dy dx = \frac{1}{6}$$
Not that, for example, $y$ doesn't actually have to be less than $1-x$ but we can suppose it is just because $x,y,z$ ~ $unif(0,1)$. It could have equivalently been
$$x \lt 1-y-z$$
$$z \lt 1-y$$
$$y \lt 1$$
You can interchange these variables as you'd like, as long as you change the order in which you integrate accordingly.