1. **State the problem:** We are testing the fairness of a coin with null hypothesis $H_0: p=0.5$ where $p$ is the probability of heads. 2. **Random variable:** $X \sim \text{Binomial}(n=6, p)$ counts heads in 6 tosses. 3. **Type I error (\alpha):** Probability of rejecting $H_0$ when it is true ($p=0.5$). 4. **Rejection Rule A:** Reject if $|X-3| \geq 3$, i.e., $X \leq 0$ or $X \geq 6$. Calculate $P(\text{Type I error for A}) = P(X=0) + P(X=6)$ under $p=0.5$: $$P(X=k) = \binom{6}{k} (0.5)^k (0.5)^{6-k} = \binom{6}{k} (0.5)^6$$ $$P(X=0) = \binom{6}{0} (0.5)^6 = 1 \times (0.5)^6 = 0.015625$$ $$P(X=6) = \binom{6}{6} (0.5)^6 = 1 \times (0.5)^6 = 0.015625$$ $$P(\text{Type I error for A}) = 0.015625 + 0.015625 = 0.03125 \approx 0.03$$ 5. **Rejection Rule B:** Reject if $X \leq 1$. Calculate $P(\text{Type I error for B}) = P(X=0) + P(X=1)$ under $p=0.5$: $$P(X=1) = \binom{6}{1} (0.5)^6 = 6 \times 0.015625 = 0.09375$$ $$P(\text{Type I error for B}) = 0.015625 + 0.09375 = 0.109375 \approx 0.11$$ 6. **Type II error (\beta):** Probability of failing to reject $H_0$ when $H_1$ is true. Given $H_1: p=0.2$. 7. **Type II error for Rule A:** Fail to reject means $|X-3| < 3$, i.e., $1 \leq X \leq 5$. Calculate $P(1 \leq X \leq 5)$ under $p=0.2$: $$P(X=k) = \binom{6}{k} (0.2)^k (0.8)^{6-k}$$ Calculate each term: $P(X=1) = \binom{6}{1} (0.2)^1 (0.8)^5 = 6 \times 0.2 \times 0.32768 = 0.39322$ $P(X=2) = \binom{6}{2} (0.2)^2 (0.8)^4 = 15 \times 0.04 \times 0.4096 = 0.24576$ $P(X=3) = \binom{6}{3} (0.2)^3 (0.8)^3 = 20 \times 0.008 \times 0.512 = 0.08192$ $P(X=4) = \binom{6}{4} (0.2)^4 (0.8)^2 = 15 \times 0.0016 \times 0.64 = 0.01536$ $P(X=5) = \binom{6}{5} (0.2)^5 (0.8)^1 = 6 \times 0.00032 \times 0.8 = 0.00154$ Sum: $$0.39322 + 0.24576 + 0.08192 + 0.01536 + 0.00154 = 0.7378 \approx 0.74$$ 8. **Type II error for Rule B:** Fail to reject means $X > 1$. Calculate $P(X > 1)$ under $p=0.2$: $$P(X > 1) = 1 - P(X \leq 1) = 1 - (P(X=0) + P(X=1))$$ Calculate $P(X=0)$: $$P(X=0) = \binom{6}{0} (0.2)^0 (0.8)^6 = 1 \times 1 \times 0.262144 = 0.26214$$ Sum $P(X \leq 1)$: $$0.26214 + 0.39322 = 0.65536$$ Therefore, $$P(X > 1) = 1 - 0.65536 = 0.34464 \approx 0.34$$ **Final answers rounded to two decimals:** - Type I error for Rule A: **0.03** - Type I error for Rule B: **0.11** - Type II error for Rule A: **0.74** - Type II error for Rule B: **0.34**