Two dice are rolled the probability that the sum of the numbers on the uppermost faces is 10 is

If two dice are rolled simultaneously, find the probability of the following events.1 The sum of the digits on the upper faces is at least 10 .2 The sum of the digits on the upper faces is 33 .3 The digit on the first die is greater than the digit on second die.

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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) = 36

 Event A: sum of the digits on the upper faces is at least 10A = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)}

∴ n(A) = 6 

`therefore P(A) = (n(A))/(n(S))`

`=6/36`

`=1/6`

Hence, the probability that the sum of the digits on the upper faces is at least 10 is`=1/6`

Page 2

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) = 36

Event B: sum of the digits on the upper faces is 33B = {}

∴ n(B) = 0 

`therefore P(A) = (n(B))/(n(S)`

`= 0/36`

  = 0

​Hence, the probability that the sum of the digits on the upper faces is 33 is 0.

Page 3

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) = 36

Event C: digit on the first die is greater than the digit on second dieC = {(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)}

∴ n(C) = 15 

`therefore P (C) =(n(C))/(n(S))`

 `=15/36`

`=5/12`

​Hence, the probability that digit on the first die is greater than the digit on second die is `=5/12`