A die is thrown twice. What is the probability that (i) 5 will not come up either time? (ii) 5 will come up at least once?

Step 1: Understand the given question and make a note of the all possible outcomes.
NOTE: A die is thrown twice.
So, Total number of event n(T) = 6 * 6 = 36
Let us consider an event A of getting number 5 on at least one dice
Events of getting 5 are n(F)= {(5,1),( 5,2),( 5,3),( 5,4),( 5,5),( 5,6),(1,5),(2,5),(3,5), (4,5),( 6,5)} = 11
Step 2: Verify that the given two events are complementary events are not.
NOTE: The given two events are compliment to each other
So, Probability of getting getting 5 = \frac{n(F)}{n(T)}
= \frac{11}{36}
Let us consider an event B of not getting number 5 either time
The probability of not getting 5 either time P(B) = 1 - P(A)
P(B) = 1 - \frac{11}{36}
= \frac{11 - 36}{36}
P(B) = \frac{25}{36}