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Answer: $\frac{4}{5}$
## Explanation We use the probability formula for the union of two events: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Where: - $A$ = event that driver is female - $B$ = event that driver lives in territory B ### Step 1: Calculate individual probabilities **Total drivers:** 300 male + 200 female = 500 drivers **$P(\text{Female})$:** $\frac{200}{500} = \frac{2}{5}$ **$P(\text{Living in territory B})$:** - Total drivers in territory A: 100 male + 100 female = 200 drivers - Total drivers in territory B: 500 - 200 = 300 drivers - $P(\text{Living in territory B}) = \frac{300}{500} = \frac{3}{5}$ **$P(\text{Female and living in territory B})$:** - Female drivers in territory A: 100 - Female drivers in territory B: 200 - 100 = 100 - $P(\text{Female and living in territory B}) = \frac{100}{500} = \frac{1}{5}$ ### Step 2: Apply the formula $$P(\text{Female or living in territory B}) = \frac{200}{500} + \frac{300}{500} - \frac{100}{500} = \frac{400}{500} = \frac{4}{5}$$ Therefore, the probability that a randomly selected driver is either female or lives in territory B is $\frac{4}{5}$.
Author: Tanishq Prabhu
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A company insures red and black cars, male and female drivers and writes policies in 2 territories A and B. There are 300 male drivers and 200 female drivers in total. There are 150 males who drive red cars and 100 females who drive red cars. 100 male and 100 female drivers live in territory A and 50 of each, males and females, drive red cars in territory A. What is the probability that a randomly selected driver is either female or lives in territory B?
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