Question

Alice and Bob arrange to meet for lunch on a certain day at noon. However, neither is known for punctuality. They both arrive independently at uniformly distributed times between noon and 1 pm on that day. Each is willing to wait up to 15 minutes for the other to show up. What is the probability they will meet for lunch that day?

Answer 1

Answer:

0.4375

Step-by-step explanation:

Lets say that X is the random varaible that determine the arrival time of Bob and Y the random variable that determine the arrival time of Alice. Bot X and Y are Independent random variables with uniform [0,1] distribution. 15 minutes is the quarter of an hour, so we want to calculate P(|X-Y|) < 0.25.

Note that P(|X-Y| < 0.25) = P(|X-Y| < 0.25 | X ≥ Y) * P(X≥ Y) + P(|X-Y| < 0.25 | Y ≥ X) * P(Y ≥ X) = P(X-Y < 0.25 | X ≥Y) * P(X≥Y) + P(Y-X < 0.25 | Y≥X)*P(Y ≥ X).

For a simmetry argument, that expression is equivalent to 2*P(Y-X < 0.25 | Y≥X)*P(Y ≥ X) = 2*P(0 < Y-X < 0.25). The region 0 < Y-X < 0.25 is, for X between 0 and 0.75, a parallelogram, of base 0.75 and height 0.25, and for X between 0.75 and 1, it is a Triangle of base and height equal to 0.25. Therefore P(0 < Y-X < 0.25) = 0.25*0.75 + 0.25² * 0.5 = 7/32. Hence 2*P(0 < Y-X < 0.25) = 14/32 = 0.4375.

They will meet for lunch with probability 0.4375