A die is thrown twice. What is the probability that:
(i) 3 will not come up either time?
(ii) 3 will come up at least once?

Answer

To solve the problem, we will break it down into two parts as stated in the question.

### Part (i): Probability that 3 will not come up either time

1. **Determine Total Outcomes**:
When a die is thrown, there are 6 possible outcomes (1, 2, 3, 4, 5, 6). When a die is thrown twice, the total number of outcomes is:
\[ 6 \times 6 = 36 \]

2. **Determine Favorable Outcomes**:
We need to find the outcomes where 3 does not appear on either die. The possible outcomes for each die, excluding 3, are (1, 2, 4, 5, 6). Therefore, there are 5 outcomes for each die. The total favorable outcomes where 3 does not appear on either die is:
\[ 5 \times 5 = 25 \]

3. **Calculate Probability**:
The probability that 3 does not come up either time is given by the ratio of favorable outcomes to total outcomes:
\[ P(\text{3 not coming up}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{25}{36} \]

### Part (ii): Probability that 3 will come up at least once

1. **Use Complement Rule**:
Instead of calculating the probability of getting a 3 directly, we can use the complement rule. The probability that 3 comes up at least once is:
\[ P(\text{at least one 3}) = 1 - P(\text{no 3s}) \]

2. **Substitute Values**:
From part (i), we found that the probability of not getting a 3 is:
\[ P(\text{no 3s}) = \frac{25}{36} \]
Therefore, the probability of getting at least one 3 is:
\[ P(\text{at least one 3}) = 1 - \frac{25}{36} = \frac{36 - 25}{36} = \frac{11}{36} \]

### Final Answers:
(i) The probability that 3 will not come up either time is \(\frac{25}{36}\).
(ii) The probability that 3 will come up at least once is \(\frac{11}{36}\).