1. **State the problem:** We have probabilities related to Steve's punctuality and weather conditions. We want to find: (b) Probability Steve is late on a rainy day. (c) Probability Steve is late. (d) Probability the day is not rainy given Steve is on time. (e)(i) Probability Steve is on time at least 3 out of 4 days. (e)(ii) Expected weekly bonus based on punctuality. 2. **Given data:** - $P(\text{Raining})=0.2$, $P(\text{Not Raining})=0.8$ - $P(\text{On time}|\text{Raining})=0.4$, $P(\text{Late}|\text{Raining})=0.6$ - $P(\text{On time}|\text{Not Raining})=0.7$, $P(\text{Late}|\text{Not Raining})=0.3$ 3. **(b) Find $P(\text{Late} \cap \text{Raining})$:** $$P(\text{Late} \cap \text{Raining}) = P(\text{Raining}) \times P(\text{Late}|\text{Raining}) = 0.2 \times 0.6 = 0.12$$ 4. **(c) Find $P(\text{Late})$ using total probability:** $$P(\text{Late}) = P(\text{Late} \cap \text{Raining}) + P(\text{Late} \cap \text{Not Raining})$$ $$= 0.2 \times 0.6 + 0.8 \times 0.3 = 0.12 + 0.24 = 0.36$$ 5. **(d) Find $P(\text{Not Raining}|\text{On time})$ using Bayes' theorem:** $$P(\text{On time}) = P(\text{On time} \cap \text{Raining}) + P(\text{On time} \cap \text{Not Raining})$$ $$= 0.2 \times 0.4 + 0.8 \times 0.7 = 0.08 + 0.56 = 0.64$$ $$P(\text{Not Raining}|\text{On time}) = \frac{P(\text{On time} \cap \text{Not Raining})}{P(\text{On time})} = \frac{0.8 \times 0.7}{0.64} = \frac{0.56}{0.64}$$ Intermediate step with cancellation: $$= \frac{\cancel{0.56}}{\cancel{0.64}} = 0.875$$ 6. **(e)(i) Probability Steve is on time at least 3 days out of 4:** Let $p = P(\text{On time}) = 0.64$ from step 5. We use the binomial distribution: $$P(X \geq 3) = P(X=3) + P(X=4)$$ $$= \binom{4}{3} p^3 (1-p)^1 + \binom{4}{4} p^4 (1-p)^0$$ Calculate each term: $$\binom{4}{3} = 4$$ $$P(X=3) = 4 \times (0.64)^3 \times (0.36) = 4 \times 0.262144 \times 0.36 = 0.3775$$ $$P(X=4) = 1 \times (0.64)^4 = 0.1678$$ Sum: $$P(X \geq 3) = 0.3775 + 0.1678 = 0.5453$$ 7. **(e)(ii) Expected weekly bonus:** - Bonus for exactly 3 days on time: 400 - Bonus for exactly 4 days on time: 1000 Calculate probabilities: $$P(X=3) = 0.3775$$ $$P(X=4) = 0.1678$$ Expected bonus: $$E = 400 \times 0.3775 + 1000 \times 0.1678 = 151 + 168 = 319$$ **Final answers:** (b) $0.12$ (c) $0.36$ (d) $0.875$ (e)(i) $0.5453$ (e)(ii) $319$ (nearest dollar)