1. **Problem Statement:** We have three independent events A, B, and C representing hits by three guns with probabilities $P(A)=0.5$, $P(B)=0.6$, and $P(C)=0.8$. We want to find: i) Probability exactly one gun hits the target. ii) Probability at least two guns hit the target. 2. **Key Formula and Rules:** Since A, B, and C are independent, the probability of combined events is the product of their probabilities. 3. **Calculations:** **i) Exactly one hit:** This means one event occurs and the other two do not. $$P(\text{exactly one}) = P(A)P(B^c)P(C^c) + P(A^c)P(B)P(C^c) + P(A^c)P(B^c)P(C)$$ Where $P(A^c) = 1 - P(A)$, etc. Calculate complements: $P(A^c) = 1 - 0.5 = 0.5$ $P(B^c) = 1 - 0.6 = 0.4$ $P(C^c) = 1 - 0.8 = 0.2$ Calculate each term: $P(A)P(B^c)P(C^c) = 0.5 \times 0.4 \times 0.2 = 0.04$ $P(A^c)P(B)P(C^c) = 0.5 \times 0.6 \times 0.2 = 0.06$ $P(A^c)P(B^c)P(C) = 0.5 \times 0.4 \times 0.8 = 0.16$ Sum: $$0.04 + 0.06 + 0.16 = 0.26$$ **ii) At least two hits:** This means either exactly two or all three hit. $$P(\text{at least two}) = P(\text{exactly two}) + P(\text{all three})$$ Calculate exactly two hits: $$P(\text{exactly two}) = P(A)P(B)P(C^c) + P(A)P(B^c)P(C) + P(A^c)P(B)P(C)$$ Calculate each term: $P(A)P(B)P(C^c) = 0.5 \times 0.6 \times 0.2 = 0.06$ $P(A)P(B^c)P(C) = 0.5 \times 0.4 \times 0.8 = 0.16$ $P(A^c)P(B)P(C) = 0.5 \times 0.6 \times 0.8 = 0.24$ Sum exactly two: $$0.06 + 0.16 + 0.24 = 0.46$$ Calculate all three hits: $$P(A)P(B)P(C) = 0.5 \times 0.6 \times 0.8 = 0.24$$ Sum for at least two: $$0.46 + 0.24 = 0.7$$ 4. **Final Answers:** - Probability exactly one hits the target: $0.26$ - Probability at least two hit the target: $0.7$