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

0.0076, if you meant 4.125 divided by 0.03125 it would be 132

step-by-step explanation:

first you have to move the decimal in 0.03125 over until its behind the 5 which is 5 places over. now the number is 3125

then you have to move the decimal in 4.125, 5 places over. now the number is 4,112,500.

now you divide like you would normally would and get 0.0076