If a die is rolled twice, the total number of outcomes is 36.
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)}
So, n(S) = 36
Let A be the event of getting 5 on either time.
The favourable outcomes are {(1,5), (2,5), (3,5), (4,5), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,5)}
So, n(A) = 11
Therefore, the probabaility of getting 5 on either time, P(A) = n(A)/n(S)
P(A) = 11/36
So, the probability that 5 will not come up on either time = 1 – P(A)
= 1 – (11/36)
= 25/36
Hence, the probability that 5 will not come up either time is 25/36.