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$X \sim R(0, 2)$, $Y \sim R(0, 5)$, X and Y are independent, I need to find $P(|X-Y| \leq 1)$

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closed as off-topic by Kavi Rama Murthy, StubbornAtom, José Carlos Santos, YuiTo Cheng, Ernie060 May 7 at 12:20

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    $\begingroup$ What does $R(a, b)$ stand for? $\endgroup$ – Yanior Weg May 7 at 11:17
  • $\begingroup$ @YaniorWeg Rectangular/Uniform distribution presumably. $\endgroup$ – StubbornAtom May 7 at 11:19
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Assuming that "R(a,b)" is the uniform distribution of a< x< b, this situation can be modeled by the rectangle in an xy-coordinate system with vertices (0, 0), (2, 0), (0,5), and (2,5). That has total area 2*5= 10. |x- y|= 1 is the same as x- y= 1 so y= x- 1, and y- x= 1 so y= x+ 1. Those are parallel lines. The region in which |x- y|< 1 is the region between those two lines. It has area 3.5 so the probability is 3.5/10= 0.35.

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