1. **Problem statement:** A coin is flipped twice and then a four-sided die (with numbers 1, 2, 3, and 4) is rolled. The coin and the die are both fair. (i) Complete the sample space. (ii) Find the probability of getting TT3. (iii) Find the probability of getting two heads followed by an even number. (iv) Find the probability of getting one head, one tail, and an odd number. --- 2. **Sample space:** Each coin flip has 2 outcomes (H or T), so two flips have $2 \times 2 = 4$ outcomes: HH, HT, TH, TT. The die has 4 outcomes: 1, 2, 3, 4. The sample space consists of all combinations of two coin flips followed by one die roll: $$\{HH1, HH2, HH3, HH4, HT1, HT2, HT3, HT4, TH1, TH2, TH3, TH4, TT1, TT2, TT3, TT4\}$$ There are $4 \times 4 = 16$ total outcomes. --- 3. **Probability of TT3:** The event TT3 is exactly one outcome in the sample space. Probability = $\frac{\text{number of favorable outcomes}}{\text{total outcomes}} = \frac{1}{16}$. --- 4. **Probability of two heads followed by an even number:** Two heads means coin flips are HH. Even numbers on the die are 2 and 4. Favorable outcomes: HH2, HH4 (2 outcomes). Probability = $\frac{2}{16} = \frac{1}{8}$. --- 5. **Probability of one head, one tail, and an odd number:** One head and one tail means coin flips are HT or TH. Odd numbers on the die are 1 and 3. Favorable outcomes: HT1, HT3, TH1, TH3 (4 outcomes). Probability = $\frac{4}{16} = \frac{1}{4}$. --- **Summary:** (i) Sample space has 16 outcomes listed above. (ii) $P(\text{TT3}) = \frac{1}{16}$. (iii) $P(\text{two heads and even number}) = \frac{1}{8}$. (iv) $P(\text{one head, one tail, odd number}) = \frac{1}{4}$.