Will they meet? Two random arrivals

Question

Two friends agree to meet between noon and 1pm. Each arrives at a time uniform on the hour, independently, and waits exactly $15$ minutes before leaving (never past 1pm).

What is the probability they meet?

Accepted Answer

Let X, Y \sim U(0, 60) (minutes past noon) be the arrival times. They meet iff X - Y \le 15. In the 60 	imes 60 square, the complement, no meeting, consists of two congruent right triangles where one arrives more than 15 minutes after the other (X - Y 15 or Y - X 15), each with legs 60 - 15 = 45: P(	ext{no meeting}) = \frac{2 \cdot \frac{1}{2}(45)^2}{60^2} = \frac{45^2}{60^2} = \left(\frac{3}{4}ight)^2 = \frac{9}{16}. Therefore P(	ext{meet}) = 1 - \frac{9}{16} = \frac{7}{16} = 0.4375. General form. With waiting time w out of a window T, P(	ext{meet}) = 1 - \left(1 - \frac{w}{T}ight)^2…

Will they meet? Two random arrivals

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