Question

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

Answer 1

$\huge{\mathcal{\color{navy}{QuesTion:}}}$

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

$\huge{\mathcal{\color{navy}{SoluTion:}}}$

A dice 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)}

n(S) = 6² = 36

Let A be the event of getting higher number in the second roll than the first.

Support of 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

The required probability assigned with event A will be :

$\frac{n(A)}{n(S)}\\= = > \frac{15}{36}\\= = > \frac{5}{12}$

♡_SítαRєddч_♡

✿_Mαrk ít αs вrαínlíєst_✿

Answer 2

$\huge{\mathfrak{\star{\underline{\underline{\red{Answer}}}}}}$

❥A dice is rolled twice. So sample space,

S = (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)}

2

n(S) = 6 = 36

Let A be the event of getting higher number in the second roll than the first.

Support of 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

The required probability assigned with event A will be :

n(A)/n(S)

15/36

5/12