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

answer: a clock

step-by-step explanation:

on friday there were 15 birds