Answer:
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
