Solution: Represent the arrival times as points $(x, y)$ in a 60-minute square. The condition $ y > x + 15 $ (order matters) corresponds to a triangular region with area $ \frac{1}{2} \times (60 - 15)^2 = \frac{1}{2} \times 2025 = 1012.5 $. The total area is $ 60^2 = 3600 $. The probability is $ \frac{1012.5}{3600} = \frac{9}{32} $.

["Understanding Probability Through Geometric Representation: The Case of Arrival Times", "When analyzing arrival times in a fixed window, mathematicians often turn to geometry to visualize and compute probabilities. This article explores a classic method: representing arrival times as points in a 60-minute × 60-minute square, where each point ((x, y)) corresponds to the arrival of someone at minute (x) and their reference timing (y). By imposing a condition such as (y > x + 15), we define a triangular region that captures a meaningful subset of all possible outcomes—revealing both geometric insight and probabilistic understanding.", "---", "### The 60-Minute Square: A Foundation for Probability", "Imagine a square with sides of length 60 minutes—each axis from (x = 0) to (x = 60) (arrival time) and (y = 0) to (y = 60) (order or reference time). Every point inside or on the boundary represents a valid pair of arrival times within the 60-minute window. The total area of this square is:\n[\n60 \ imes 60 = 3600 \ ext{ square minutes}.\n]", "---", "### Defining the Condition ( y > x + 15 )", "Now, consider the line ( y = x + 15 ). This line rises diagonally from ((0, 15)) to ((45, 60)) on the square—cutting off a triangular region where arrivals are earlier than allowed by the condition. Specifically, ( y > x + 15 ) means the second person arrives significantly later than the first plus 15 minutes.", "The region satisfying this inequality forms a right triangle bounded by:\n- The line ( y = x + 15 ),\n- The top horizontal line ( y = 60 ),\n- And the vertical line ( x = 45 ).", "The triangle’s base runs from (x = 0) to (x = 45), and its height at (x = 0) is (60 - 15 = 45). Alternatively, since both x and y values max at 60, the base and height of the triangle are both (60 - 15 = 45).", "---", "### Calculating the Area of the Triangular Region", "The area of a triangle is given by:\n[\n\ ext{Area} = \frac{1}{2} \ imes \ ext{base} \ imes \ ext{height} = \frac{1}{2} \ imes 45 \ imes 45 = \frac{1}{2} \ imes 2025 = 1012.5 \ ext{ square units}.\n]", "This area represents the favorable outcomes where the second arrival time occurs more than 15 minutes after the first.", "---", "### Computing the Probability", "To find the probability of this condition, divide the area of the favorable region by the total area:\n[\nP(y > x + 15) = \frac{1012.5}{3600}.\n]", "Simplify the fraction:\n[\n\frac{1012.5}{3600} = \frac{10125}{36000} = \frac{9}{32}.\n]", "Thus, the probability that the second arrival is more than 15 minutes after the first is exactly ( \frac{9}{32} ), or approximately 28.125%.", "---", "### Why This Geometric Approach Works", "By embedding probabilistic conditions in a geometric framework, we transform abstract numbers into visual, intuitive shapes. This method simplifies complex computation—like finding areas bounded by linear inequalities—and offers immediate insight into how constraints shape possible outcomes. It’s a powerful tool in probability, statistics, and risk analysis.", "---", "Key Takeaway:\nRepresenting arrival times as coordinates and using geometric rules to compute regions under inequalities is a clean, efficient way to analyze probability problems. For the condition (y > x + 15), we obtain a triangle with area 1012.5 and probability ( \frac{9}{32} )—a result both precise and visually compelling.", "---", "Keywords: probability geometry, arrival time analysis, geometric probability, triangular region area, order-based constraints, 60-minute square, calculation probability (\frac{9}{32})"]