A single die is rolled twice. The 36 equally-likely outcomes are shown to the right. Find the probability of getting A second number that is less than the first number

The probability of getting A second number that is less than the first number is

Step-by-step explanation:

Step(i):-

Given a single die is rolled twice

In throwing a die , there are six exhaustive elementary  events

1 or 2 or 3 or 4 or 5 or 6.

The total number of exhaustive events = 6

Given data a single die is rolled two times = 6² = 36

The total number of exhaustive cases n(S) = 36

Step(ii):-

Let 'E' be the event of getting A second number that is less than the first number.

The required pairs are

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

The total number of favorable cases n(E) = 15

The required probability

Conclusion:-

The probability of getting A second number that is less than the first number is

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

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step-by-step explanation:

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