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Answer: 45%
## Explanation We use the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Where: - \( P(A \cup B) = 0.60 \) (probability of qualifying for discount) - \( P(A) = 0.15 \) (probability of having sprinkler system) - \( P(B) = x \) (probability of living within 5 miles of fire station - what we're solving for) - \( P(A \cap B) = 0 \) (since events are mutually exclusive) Since the events are mutually exclusive, \( P(A \cap B) = 0 \), so the formula simplifies to: \[ P(A \cup B) = P(A) + P(B) \] \[ 0.60 = 0.15 + x \] \[ x = 0.60 - 0.15 \] \[ x = 0.45 = 45\% \] Therefore, the probability that a randomly selected homeowner lives within 5 miles of a fire station is **45%**.
Author: Tanishq Prabhu
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A homeowners insurer offers a discount for homeowners that either have a sprinkler system or live within 5 miles of a fire station. 60% of homeowners qualify for the discount. Only 15% of homeowners have a sprinkler system, and none of those homeowners live within 5 miles of a fire station. If the events are mutually exclusive, what is the probability a randomly selected homeowner lives within 5 miles of a fire station?
A
15%
B
25%
C
30%
D
45%