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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?

 Jan 7, 2021
 #1
avatar+303 
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7/24

 Jan 7, 2021
 #2
avatar+218 
+1

Please explain. I'm looking for how to do it.

 Jan 7, 2021
 #3
avatar+303 
-1

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: \(\frac{8/3-1/6-1/6}8=\frac{7/3}8=\boxed{7/24}\)

 Jan 7, 2021
 #4
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+1

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

 Jan 12, 2021

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