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

your question is incomplete, write all the numbers in the sequence that it tells you.  

step-by-step explanation: