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A single die is rolled twice in succession. What is the probability that the number on the second roll is greater than that on the first roll?

Given, A die is rolled twice,

so, sample space 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)}

number of samples n(S) = 6² = 36

Let A be the event of getting a greater number in the second roll than that in the first roll

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

n(A) = 15

Therefore, the required probability of event A =$\frac{n\left(A\right)}{n\left(S\right)}$=$\frac{15}{36}=\frac{5}{12}$