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Answer: 76.5%
These events are independent, so $$P(A' \text{ and } B') = P(A') \times P(B') = (1 - 0.15) \times (1 - 0.10) = 0.765 \text{ or } 76.5\%$$ **Explanation:** - Let A be the event that the male driver has a claim: P(A) = 0.15 - Let B be the event that the female driver has a claim: P(B) = 0.10 - The probability that the male driver does NOT have a claim: P(A') = 1 - 0.15 = 0.85 - The probability that the female driver does NOT have a claim: P(B') = 1 - 0.10 = 0.90 - Since the events are independent (the claim status of one driver doesn't affect the other), the probability that neither has a claim is: P(A' and B') = P(A') × P(B') = 0.85 × 0.90 = 0.765 = 76.5%
Author: Nikitesh Somanthe
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A company insures male and female drivers. Males have .15 chance of having a claim during a policy period and females have .10 chance of having a claim. If a male and female driver is randomly selected from the population, what is the probability neither one has a claim during the policy period?
A
73.5%
B
76.5%
C
77.5%
D
85%
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