Question

A fair six-sided die (with outcomes 1 through 6) is rolled twice, and two numbers are obtained. X1 = result of the first roll, and X2 = result of the second roll. Let A be the event that X1 > X2. What is the probability of A?

Answer 1

Answer:

$\dfrac{5}{12}$

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.

$Probability=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

$Probability=\dfrac{15}{36}$

$Probability=\dfrac{5}{12}$

Therefore, the probability of A is $\dfrac{5}{12}$.