Ok so,
Let the number of hours after 2:00 p.m. that the two engineers arrive at Starbucks be x and y, and let the number of hours after 2:00 p.m. that the boss arrives at Starbucks be z. Then \(0\le x,y,z\le2\) and in three dimensions, we are choosing a random point within this cube with volume 8. We must have \(z>x\) and \(z>y\); this forms a square pyramid with base area 4 and height 2, or volume 8/3.
However, if one of the engineers decides to leave early, the meeting will fail. The engineers will leave early if \(x>y+1\) or \(y>x+1\). The intersections of these with our pyramid gives two smaller triangular pyramids each with base area 1/2 and height 1, or volume 1/6.
In all, the probability of the meeting occurring is the volume of the big square pyramid minus the volumes of the smaller triangular pyramids divided by the volume of the cube: \(\frac{8/3-1/6-1/6}8=\frac{7/3}8=\boxed{7/24}\)
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