1. **State the problem:** We have a basketball player with the following probabilities: - Probability of making the first free throw: $P(M_1) = 0.8$ - Probability of missing the first free throw: $P(M_1^c) = 0.2$ - Probability of making the second free throw given the first was missed: $P(M_2|M_1^c) = 0.7$ - Probability of making the second free throw given the first was made: $P(M_2|M_1) = 0.8$ (assuming same success rate as first shot if first is made) We want to find: - Probability the player makes both shots: $P(M_1 \cap M_2)$ - Probability the player makes exactly one free throw - Probability the player missed the first free throw given they make the second one: $P(M_1^c|M_2)$ 2. **Find the probability the player makes both shots:** Using the multiplication rule for dependent events: $$P(M_1 \cap M_2) = P(M_1) \times P(M_2|M_1)$$ Substitute values: $$P(M_1 \cap M_2) = 0.8 \times 0.8 = 0.64$$ 3. **Find the probability the player makes exactly one free throw:** This can happen in two ways: - Make first, miss second: $P(M_1 \cap M_2^c)$ - Miss first, make second: $P(M_1^c \cap M_2)$ Calculate each: $$P(M_1 \cap M_2^c) = P(M_1) \times P(M_2^c|M_1) = 0.8 \times (1 - 0.8) = 0.8 \times 0.2 = 0.16$$ $$P(M_1^c \cap M_2) = P(M_1^c) \times P(M_2|M_1^c) = 0.2 \times 0.7 = 0.14$$ Add them: $$P(\text{exactly one}) = 0.16 + 0.14 = 0.3$$ 4. **Find the probability the player missed the first free throw given they make the second one:** Use Bayes' theorem: $$P(M_1^c|M_2) = \frac{P(M_1^c \cap M_2)}{P(M_2)}$$ Calculate $P(M_2)$ using total probability: $$P(M_2) = P(M_1) \times P(M_2|M_1) + P(M_1^c) \times P(M_2|M_1^c) = 0.8 \times 0.8 + 0.2 \times 0.7 = 0.64 + 0.14 = 0.78$$ Now substitute: $$P(M_1^c|M_2) = \frac{0.14}{0.78} \approx 0.1795$$ **Final answers:** - Probability both shots made: $0.64$ - Probability exactly one shot made: $0.3$ - Probability missed first given made second: $0.1795$