1. **Problem statement:** We toss a fair coin 3 times. (1) Find the probability of exactly 2 heads. (2) Find the probability of more than 1 tail. (3) Find the probability of all heads. --- 2. **Understanding the problem:** Each coin toss has 2 possible outcomes: Heads (H) or Tails (T). Total number of possible outcomes in 3 tosses is $$2^3 = 8$$. 3. **Step 1: Probability of exactly 2 heads.** Number of ways to get exactly 2 heads out of 3 tosses is the binomial coefficient $$\binom{3}{2}$$. Calculate: $$\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3$$ Probability is number of favorable outcomes over total outcomes: $$P(\text{exactly 2 heads}) = \frac{3}{8}$$ 4. **Step 2: Probability of more than 1 tail.** More than 1 tail means 2 or 3 tails. Calculate ways: - Exactly 2 tails: $$\binom{3}{2} = 3$$ - Exactly 3 tails: $$\binom{3}{3} = 1$$ Total favorable outcomes = $$3 + 1 = 4$$ Probability: $$P(\text{more than 1 tail}) = \frac{4}{8} = \frac{1}{2}$$ 5. **Step 3: Probability of all heads.** There's only 1 way to get all heads: HHH. Probability: $$P(\text{all heads}) = \frac{1}{8}$$ --- **Final answers:** (1) $$\frac{3}{8}$$ (2) $$\frac{1}{2}$$ (3) $$\frac{1}{8}$$