Alice and Bob need to discuss a business project for 5 minutes. However, they haven't coordinated when they are going to be in the office, and in fact they are going to follow strict schedules of being in the office for only 15 minutes each. Each will arrive at an independently chosen random time between 2 PM and 3 PM, and their 5 minute discussion must end by 3 PM.
Given the randomness and lack of coordination, what is the probability that Alice and Bob will get to complete their 5-minute discussion?
As the times that Alice and Bob are in the office must overlap by at least 5 minutes, and are 15 minutes long, if we know one time, the other must be within 10 minutes or their meeting will be shorter than 5 minutes.
Now, from 2PM to 2:10PM, the left side 10 minutes is shortened to just m minutes, where m is the number of minutes past the hour. So, from 2PM to 2:10PM, we have a 10 + m minute range, or a probability of lining up at a given time of, (10 + m) / 60. As I chose to work in minutes, we must integrate (1 / 60) * (10 + m) / 60 from m = 0 to 10.
1/3600*(10m + (m^2)/2) for m = 10 becomes ... 1/3600 * (10*10 + 10^2 / 2) = 1/3600 * (100 + 50) = 150 / 3600 = 1 / 24.
Looking at the upper end, it is clear that both times must be before 2:55PM. From 2:45PM to 2:55PM, we have the same as above, where the left side range is 10 minutes, but the right side range is smaller (in this case 55-m). Integrating that returns the same result as above. So for the two ends, we have a probability of 2/24 or 1/12.
Now, from 2:10PM to 2:45PM, we have the full 10 minutes on either side, so a total of 20/60 or 1/3 for each time. 1/3 * 35 / 60 = 35 / 180 = 7 / 36.
1 / 12 + 7 / 36 = 3 / 36 + 7 / 36 = 10 / 36 or 5 / 18.
Thus, we should have a 5 / 18 probability that Alice and Bob are able to complete their 5-minute meeting.
Posted by Justin
on 2011-02-04 14:21:46