The element of A is 1 whereas the elements of A’ are 2, 3, 4, 5, 6
As you can recall n(A) + n(A’) = n(U)
Adopting this into probability, the probability of getting the smallest number when a die is rolled: 1/6
The probability of getting the other numbers when a die is
rolled: 5/6
So for the event of A, we have P(A)+P(A’) = 1/6 + 5/6 = 1
So we can conclude that the possibility of an event A not occurring can be evaluated as; P(A') = 1 - P(A) 