Answer:
[tex]\dfrac{5}{12}[/tex]
Step-by-step explanation:
It is given that a fair six-sided die is rolled twice.
X₁= result of the first roll
X₂=result of the second roll
Let A be the event that X₁>X₂.
A={(2,1),(3,1),(4,1),(5,1),(6,1),(3,2),(4,2),(5,2),(6,2),(4,3),(5,3),(6,3),(5,4),(6,4),(6,5)} = 15
It a fair six-sided die is rolled twice then the total number of outcomes is 36.
We need to find the probability of A.
[tex]Probability=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}[/tex]
[tex]Probability=\dfrac{15}{36}[/tex]
[tex]Probability=\dfrac{5}{12}[/tex]
Therefore, the probability of A is [tex]\dfrac{5}{12}[/tex].
