Respuesta :
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
[tex]Probability=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
[tex]Probability=\dfrac{15}{36}[/tex]
[tex]Probability=\dfrac{5}{12}[/tex]
Therefore, the required probability is [tex]\dfrac{5}{12}[/tex].
