x and y are uniformly distributed over the interval [0,1]. Find the probability that |x-y|, the distance between x and y, is less than 0.4

0.72

Step-by-step explanation:

Given:

- x and y are uniformly distributed over the interval [0,1].

- |x-y|, the distance between x and y, is less than 0.4

Find:

Find the probability when |x-y| < 0.4

Solution:

- The constrained area is the portion of the unit square between the lines:

y=x-0.4 and y=x+0.4.

- That's the R2 interval:

⟨x, y⟩ ∈ [0; 1] * [max{ 0, x-0.4 }; min{ 1, x+0.4 }]

- This can be subdivided into:

([ 0; 0.4) x [ 0; x + 0.4)

⟨x, y⟩∈ (U [0.4; 0.6) * [x-0.4; x+0.4))

(U [0.6; 1) * [x-0.4; 1))

- The area enclosed is two equal units of triangles and one square. Hence, we calculate the areas:

Area of triangle = 0.5*B*H

Area of triangle = 0.5*0.8*0.8 = 0.32

Area of parallelogram = 0.4*0.2 = 0.08

- Hence probability is:

Total Area = 2*0.32 + 0.08 = 0.72