Step 1: Understand the experiment.
A die is thrown twice. Each throw has 6 possible outcomes. Therefore, the total number of possible outcomes when a die is thrown twice is:
\(6 \times 6 = 36\).
Thus, the total sample space contains 36 equally likely outcomes.
Part (i): Probability that 6 will not come up either time.
Step 2: Determine outcomes where 6 does not appear.
If 6 does not appear on a throw, the possible outcomes are:
\(1, 2, 3, 4, 5\).
So there are 5 possible outcomes for each throw that are not 6.
Step 3: Calculate favourable outcomes.
For two throws, the number of outcomes where 6 does not appear on both throws is:
\(5 \times 5 = 25\).
Step 4: Find the probability.
Probability \(=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\)
\[
P(\text{no 6 in both throws}) = \frac{25}{36}
\]
Part (ii): Probability that 6 will come up at least once.
Step 5: Use the complement rule.
“At least once” means one or two times. It is easier to calculate using the complement rule:
\[
P(\text{at least one 6}) = 1 - P(\text{no 6})
\]
Step 6: Substitute the value.
\[
P(\text{at least one 6}) = 1 - \frac{25}{36}
\]
\[
= \frac{36}{36} - \frac{25}{36}
\]
\[
= \frac{11}{36}
\]
Final Answers:
(i) Probability that 6 will not come up either time = \(\frac{25}{36}\)
(ii) Probability that 6 will come up at least once = \(\frac{11}{36}\)