SOLUTION: Consider the random experiment of tossing two fair dice and recording the up faces. Let X be the sum of the two dice, and let Y be the absolute value of the difference of the two d

There are 36 way a pair of dice can fall:

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
 
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
 
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6) 
 
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6) 
 
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
 
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

for x=0, the rolls with absolute difference 0 are these 6: 
(1,1), (2,2), (3,3), (4,4), (5,5), (6,6).  so P(0) = 6/36 = 1/6. 

for x=1, the rolls with absolute difference 1 are these 10: 
(1,2), (2,3), (3,4), (4,5), (5,6), (2,1), (3,2), (4,3), (5,3), (6,5).  
So P(1) = 10/36 = 5/18.

for x=2, the rolls with absolute difference 2 are these 8: 
(1,3), (2,4), (3,5), (4,6), (6,4), (5,3), (4,2), (3,1).  
So P(2) = 10/36 = 5/18.

for x=3, the rolls with absolute difference 3 are these 6: 
(1,4), (2,5), (3,6), (4,1), (5,2), (4,1).  
So P(3) = 6/36 = 1/6.

for x=4, the rolls with absolute difference 4 are these 4: 
(1,5), (2,6), (5,1), (6,2).  
So P(4) = 4/36 = 1/9.

for x=5, the rolls with absolute difference 5 are these 2: 
(1,6), (1,6).  
So P(5) = 2/36 = 1/18.

So the probability function of Y is 

  x   P(x) 
  0   6/36 = 1/6 
  1  10/36 = 5/18
  2   8/36 = 2/9
  3   6/36 = 1/6
  4   4/36 = 1/9
  5   2/36 = 1/18
-----------------
     36/36 = 1

The cumulative distribution function is found by accumulating
value plus the sum of the values before it.
  x   CP(x) 
  0   6/36 = 1/6 
  1   6/36+10/36 = 16/36 = 4/9 
  2   8/36+6/36+8/36 = 22/36 = 11/18
  3   6/36+8/36+6/36+6/36 = 26/39 = 2/3
  4   6/36+8/36+6/36+10/36+4/36 = 34/36 = 17/18
  5   6/36+8/36+6/36+10/36+4/36+2/36 = 36/36 = 1
-----------------
     36/36 = 1

Edwin