1. **State the problem:** A coin is tossed three times, producing 8 possible outcomes: HHT, HTH, TTT, THH, HTT, TTH, THT, HHH. We analyze three events and find their probabilities. 2. **Total number of outcomes:** Since each toss has 2 possible results, total outcomes = $2^3 = 8$. --- ### (a) Event A: Alternating head and tail (either starting with head or tail) 3. **Identify outcomes with alternating heads and tails:** These are sequences where no two adjacent tosses are the same. Check each: - HHT: H-H (not alternating), discard - HTH: H-T-H (alternating), include - TTT: T-T (not alternating), discard - THH: T-H-H (not alternating), discard - HTT: H-T-T (not alternating), discard - TTH: T-T-H (not alternating), discard - THT: T-H-T (alternating), include - HHH: H-H (not alternating), discard 4. **Event A outcomes:** $\{\text{HTH}, \text{THT}\}$ 5. **Probability of Event A:** $$P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{8} = \frac{1}{4}$$ --- ### (b) Event B: More heads than tails 6. **Count heads in each outcome:** - HHT: 2 heads, 1 tail (include) - HTH: 2 heads, 1 tail (include) - TTT: 0 heads, 3 tails (exclude) - THH: 2 heads, 1 tail (include) - HTT: 1 head, 2 tails (exclude) - TTH: 1 head, 2 tails (exclude) - THT: 1 head, 2 tails (exclude) - HHH: 3 heads, 0 tails (include) 7. **Event B outcomes:** $\{\text{HHT}, \text{HTH}, \text{THH}, \text{HHH}\}$ 8. **Probability of Event B:** $$P(B) = \frac{4}{8} = \frac{1}{2}$$ --- ### (c) Event C: A tail on both the first and last tosses 9. **Check outcomes with tail on first and last toss:** - HHT: first H (exclude) - HTH: first H (exclude) - TTT: first T, last T (include) - THH: first T, last H (exclude) - HTT: first H (exclude) - TTH: first T, last H (exclude) - THT: first T, last T (include) - HHH: first H (exclude) 10. **Event C outcomes:** $\{\text{TTT}, \text{THT}\}$ 11. **Probability of Event C:** $$P(C) = \frac{2}{8} = \frac{1}{4}$$