Another Problem Begging to Be Extended
Like a previous problem, the following problem was published in Oregon Mathematics Teacher, a great source for problems that I have used regularly with students for more than 30 years. Again the solution is nice, but the problem itself is begging to be extended...
Initial Problem: Two people agree to meet for lunch between 12:00 noon and 1:00 PM. If each agrees to wait 15 minutes for the other, what is the probability that the two will NOT have lunch together?
Once you have solved the above problem, consider this extension (one of many that seem ready to appear!).
Extension:
Two people agree to meet for lunch between 12:00 noon and 1:00 PM. If each agrees to wait N minutes for the other, what value of N leads to a probability of 0.5 that the two will NOT have lunch together? What about a probability of 0.25? A probability of M, for 0 < M <1?
Solve this new problem...And then think about an additional extension...suppose the wait time for each person is different from the other...or suppose three people were involved, making the same agreement of 15 minutes...then...
Source: Oregon Mathematics Teacher, Jan/Feb 2004, pp. 43, 45.
Hint: Play with the problem...try some times, trying to be somewhat systematic. For example, if one person arrives at 12:30, when the other person have to arrive in order for them to NOT meet?
Solution Commentary: The answer is so nice, when viewed in terms of this graph:
Let w_{1} = w_{2} = .15 (or 15 minutes)and t_{1} = t_{2} = 1 (or 1 hour).
The shaded area represents times when they will meet and have lunch. The two unshaded triangles represent NOT meeting.
Using the ideas of geometric probability, the area of each triangle is (1/2)(3/4)(3/4) = 9/32 and the area of the square is 1x1. Thus the probability of NOT meeting is [(2)(9/32)]/1 = 9/16 = 0.5625.
Now, how can you adjust it for the extension, where the wait time is N minutes for each?
