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 A pair of fair dice is rolled.

36 possible rolls       Six would  be same numbers     6/36 = 1/6

    All the other rolls would be DIFFERENT numbers      30/36   =  5/6

Subtract the = rolls...that leaves 30 rolls      1/2 of these would be second die lower      15/36  = 5/12

A) Let's say the number on the first die is X, the second one has 1/6 of getting X

B) Let's say the first number is Y, the second has 5/6 chance of not getting Y

C) Casework

   1) 1st number is 6 second is 1, 2, 3, 4, 5 chance 1/6 times 5/6 = 5/36

   2) 1st number is 5 second is 1, 2, 3, 4 chance 1/6 times 4/6 = 4/36

   3) 1st number is 4 second is 1, 2, 3 chance 1/6 times 3/6 = 3/36

   4) 1st number is 3 second is 1, 2 chance 1/6 times 2/6 = 2/36

   5) 1st number is 2 second is 1 chance 1/6 times 1/6 = 1/36

Add them up = 15/36 = 5/12

(a)  P(showing  the  same  number)   =   6  /36   =   1 / 6

(b)  P (showing different  numbers)  =  

1 - P (that they show  the same numbers)  =

1 -  1/6   =

5/6

(c)  Probability  (second die  shows lesser  number than the first) =  15/36   =  5 / 12