A six-sided die is rolled two times. What is the theoretical fractional probability that the first number is greater than the second number?

Given:

A six-sided die is rolled two times.

To find:

The theoretical fractional probability that the first number is greater than the second number.

Solution:

If a six-sided die is rolled two times, then the total possible outcomes are 36.

S = {(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 outcomes where the first number is greater than the second number are

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

So, the favorable outcomes = 15.

Now, the theoretical fractional probability that the first number is greater than the second number is

Therefore, the required probability is .

the square root of what number? if you add that, then someone will probably answer the question.

d. it is a convex pentagon because it has five sides and none of the sides would extend into the inside of the polygon.

step-by-step explanation: