A random sum near the middle

Question

Two numbers X and Y are chosen independently and uniformly at random from [0, 1]. What is the probability that 	frac12 \le X + Y \le 	frac32?

Accepted Answer

In the unit square, exclude the two corner triangles where the sum is out of range. - X + Y < 	frac12 is a corner triangle at the origin with legs 	frac12, area 	frac12 \left(	frac12ight)^2 = 	frac18. - X + Y 	frac32 is a corner triangle at (1,1) with legs 2 - 	frac32 = 	frac12, area 	frac18. So P\left(	frac12 \le X + Y \le 	frac32ight) = 1 - 	frac18 - 	frac18 = \frac34 . Takeaway. A band on the sum removes a corner triangle at each end of the square's diagonal. By symmetry the two triangles are equal here, leaving three quarters of the area.

A random sum near the middle

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