a standard six-sided die is rolled. what is the probability that the value of the second die is higher than the first

The probability that the value of the second die is higher than the first

Step-by-step explanation:

Explanation:-

In a single throw with two dice total number of sums = 6 X 6 = 36

Let 'E' be the event of the second die is higher than the first

Total number of cases

= {( 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)}

The favourable cases

      n(E) = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)} = 15

The probability that the value of the second die is higher than the first

step-by-step explanation:

if you meant, "on the way from (3, 2) to (3, 14):

x does not change; y increases by 12.

ensure that you have copied down this problem correctly.   .