The key move is recognizing that |X − Y| follows a triangular distribution on [0, 100] with PDF f(d) = 2(100 − d) / 10000.
Model the problem geometrically by considering the unit square [0,100]x[0,100] and identifying the region where 50 ≤ |X−Y| ≤ 80. Compute the area of this region and divide by 10000 to get the probability.
Pro tip: After computing the probability, sanity-check it by considering the extreme cases: P(|X−Y| ≤ 50) should be 0.75 (since the area of the band |x−y| ≤ 50 is 7500), and P(|X−Y| ≤ 80) should be 0.96 (area 9600). The difference 0.21 is plausible.
Draw the 100x100 square representing all possible outcomes (X,Y). The total area is 10000.
The condition 50 ≤ |X−Y| ≤ 80 corresponds to the area between the lines y = x ± 50 and y = x ± 80, within the square.
Calculate the area of the band where |X−Y| ≤ 80 and subtract the area where |X−Y| ≤ 50. Alternatively, compute the two triangular strips directly.
Divide the area of the region by the total area (10000) to obtain the probability.
Check the result for plausibility (e.g., between 0 and 1) and present the final probability, e.g., 0.21 or 21%.
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