Originally Posted by
mr fantastic
Let X be the random variable number of minutes after noon that person X arrives. X ~ U(0, 60).
Let Y be the random variable number of minutes after noon that person Y arrives. Y ~ U(0, 60).
1. When $\displaystyle 0 \leq X \leq 15$ you require $\displaystyle 0 \leq Y \leq X + 15$ for the two people to meet.
2. When $\displaystyle 15 \leq X \leq 45$ you require $\displaystyle X - 15 \leq Y \leq X + 15$ for the two people to meet.
3. When $\displaystyle 45 \leq X \leq 60$ you require $\displaystyle X - 15 \leq Y \leq 60$ for the two people to meet.
On the XY-plane draw the square bounded by the lines X = 0, Y = 0, X = 60 and Y = 60. The area of this square is 3600 square minutes.
Cases 1-3 above define areas inside this square (unit is square minutes) for which the two people will meet:
1. 675/2
2. 900
3. 675/2
Total area = 1350 square minutes.
Therefore the probability of the two people meeting is 1350/3600 = 7/16.
Note: This geometric approach only works because X and Y follow uniform distributions.