Answer: 45%

## Explanation We are given: - Let A = event that a homeowner has a sprinkler system - Let B = event that a homeowner lives within 5 miles of a fire station - P(A ∪ B) = 0.60 (60% qualify for the discount) - P(A) = 0.15 (15% have a sprinkler system) - P(A ∩ B) = 0 (none of those with sprinkler systems live within 5 miles of a fire station) Using the probability addition rule: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Substituting the known values: $$0.60 = 0.15 + P(B) - 0$$ $$P(B) = 0.60 - 0.15 = 0.45$$ Therefore, the probability that a randomly selected homeowner lives within 5 miles of a fire station is **45%**. **Key points:** - The problem uses the addition rule for probability - The events are mutually exclusive for those with sprinkler systems (P(A ∩ B) = 0) - The solution involves simple algebraic rearrangement

Author: Nikitesh Somanthe