Random points in a cube. 
A point with coordinates $x$,$y$,$z$, is chosen uniformly at random from a cube:
$$\{(x,y,z)\in \mathbb{R^3}:0\le x,y,z \le 10\}.$$ 
Assume that the probability of an event is proportional to the volume of a cube. What is the probability of the following events:
1) $\max(x,y,z)=3,$
2) $x+y \le 10$?

I feel for the second part we can compute it using volume of pyramid volume of cube which turns out to be $\frac 1 6$ but I'm not sure. Any help would be appreciated. Thank you.

For $X+Y\le 10$, note that $Z$ is perfectly free to roam. Thus our solid is not a pyramid, it is a prism.  Draw the part of the $10\times 10$  square in the $x$-$y$ plane that has $x+y\le 10$. This has area $50$. So the volume of our prism is $500$, half the total volume of the cube.
The first question is trickier, since we are asking for the largest of $X,Y,Z$ to be exactly $3$. That region has $0$ volume. This is because the part of our region with $x=3$ is a $10\times 10$ square. It has $0$ thickness, and therefore $0$ volume. 
The same is true of  the region with $y=3$, and the region with $z=3$. Our region  is a subset of the union of these three squares, so it too has $0$ volume. 
Remark: Note that the answer to the first question changes if we change the condition to $\max(X,Y,Z)\le 3$. For then our region is a $3\times 3\times 3$ cube. 
