Basic probability problem! 
"A boss plans a business meeting at Starbucks with the two engineers below him. However, he fails to set a time, and all three arrive at Starbucks at a random time between $2:00$ and $4:00$ p.m. When the boss shows up, if both engineers are not already there, he storms out and cancels the meeting. Each engineer is willing to stay at Starbucks alone for an hour, but if the other engineer has not arrived by that time, he will leave. What is the probability that the meeting takes place?"

What answer did you get?
Thanks!
 A: Without the "waiting for an hour" caveat, in a $2\times2\times2$ cube, you are asked to find the proportion of volume with $z\geq x,y$. If you can picture this shape, it's a pyramid with a square base with a right angle at one corner. Two of its walls are right triangles meeting at the right angle, and the other two walls are right triangles at a $45^\circ$ incline.
Adding on the "waiting for an hour" caveat, you also require $\left|x-y\right|<1$, which shaves off two smaller pyramids from this shape, each with a triangular base.
If you can picture the problem this way, then you can use simple geometry formulas to arrive at a result without integration. The volume of a pyramid whose base is $B$ and height is $h$ is $\frac{1}{3}Bh$.

Here is a picture:


I leave the explicit calculations alone since this may be homework.
A: DISCLAIMER: Please note that this is meant to be a sketch of how you would do a problem like this. I could've made a computational mistake along the way.
We have a continuous probability distribution function. The probability a given engineer is there at $2$ or $5$ is $0$ and it is equally likely that he'll be there at any time between $3$ and $4$. We know that the area under this function must be normalized and so $k=1/2$ where $k$ is the probability that an engineer will be there at any point in time in between $3$ and $4$. Why? So, we have:
$$P(\text{engineer 1 is at Starbucks at time $t$}) = \begin{Bmatrix}
\frac{t-2}{2}, 2 \leq t \leq 3\\ 
\frac{1}{2}, 3 \leq t \leq 4 \\ 
\frac{5-t}{2} 4 \leq t \leq 5 \end{Bmatrix}.$$
So, $$P(\text{engineer 1 is at Starbucks when engineer 2 arrives})  =  \begin{Bmatrix}
\int_{2}^{t} \frac{t-2}{2} = 1/4\cdot t^2 - t + 1, 2 \leq t \leq 3\\ 
\int_{t-1}^{3} \frac{t-2}{2} + \int_{3}^{t} \frac{1}{2} = \frac{3}{4} - (t-1)^2 + t - 1 + \frac{t-3}{2} = -t^2 + 7/2\cdot t - 11/4, 3 \leq t \leq 4 \\ 
\int_{t-1}^{4} \frac{1}{2} + \int_{4}^{t} \frac{5-t}{2} = 1/2\cdot (5-t) + 5/2\cdot t - 1/4\cdot t^2 - 6 = -1/4\cdot t^2 + 2\cdot t - 7/2 , 4 \leq t \leq 5 \end{Bmatrix}.$$
Now, $$P(\text{boss arrives after engineer 2}) = \begin{Bmatrix}
1-\frac{(t-2)^2}{4}, 2 \leq t \leq 3\\ 
3/4-\frac{(t-3)}{2}, 3 \leq t \leq 4 \\ 
\frac{(5-t)^2}{4}, 4 \leq t \leq 5 \end{Bmatrix}.$$
Now, you want to evaluate $P(\text{engineer 1 at Starbucks when engineer 2 arrives})\cdot P(\text{boss arrives after engineer 2})$. Then integrate over $t \in [2,5]$ and you should get the probability desired. However, this is by no means a trivial problem; I would not call it 'basic'!
A: If the engineers arrive in the first hour and the boss arrives in the second hour (p=1/8), then the meeting will definitely be held (p=1).
If all three arrive in the same hour (first or second) (p=1/8+1/8=1/4), then the meeting will be held only if the boss (who may arrive first, second, or last) arrives last (p=1/3)
If one of the engineers arrives in the first hour while the other engineer and the boss arrive in the second hour (p=1/4), then the meeting will be held only if the second engineer arrives before the boss and before the first engineer departs (the order of the arrival of the boss and the departure of the first engineer does not matter) (p=1/3)
Adding these probabilities, the total probability of the meeting taking place = (1/8)1+(1/4)(1/3)+(1/4)*(1/3) = 7/24
A: I think the solution from alex.jordan should be accepted, but I just give an advice to the OP.
When you have such a problem, you can easily verify your result with a small program, that simulates the experience a large number of times, a compute the proportion of successes. For example (this is python):
import random
def f() :
 x=random.uniform(0,2)
 y=random.uniform(0,2)
 z=random.uniform(0,2)
 if z>x and z> y and (x-y)<1 and (y-x)<1 : return 1
 return 0
a=0
for i in range(10000000) : a+=f()
print(a/10000000)

You can obtain something like : 0.2918696 and you know this must be really close to the right answer. So this can't be for example $\frac{27}{256}$
