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Let $W$ be the event that a randomly chosen individual is a woman. Let $O$ be the event that a randomly chosen individual works outside the village.
We are given the following information: Total number of people in the village = 8000 Number of people who go out of the village to work = 3000 Number of women = 4000 Percentage of women who go out of the village to work = 30%
The probability that a randomly chosen individual is a woman is: $$P(W) = \frac{\text{Number of women}}{\text{Total number of people}} = \frac{4000}{8000} = \frac{1}{2}$$
The probability that a randomly chosen individual works outside the village is: $$P(O) = \frac{\text{Number of people working outside}}{\text{Total number of people}} = \frac{3000}{8000} = \frac{3}{8}$$
We are given that 30% of women go out of the village to work. So, the number of women who work outside the village is: $$0.30 \times 4000 = 1200$$ The probability that a randomly chosen individual is a woman and works outside the village is: $$P(W \cap O) = \frac{\text{Number of women working outside}}{\text{Total number of people}} = \frac{1200}{8000} = \frac{3}{20}$$
We want to find the probability that a randomly chosen individual is either a woman or a person working outside the village, which is $P(W \cup O)$. Using the formula for the union of two events: $$P(W \cup O) = P(W) + P(O) - P(W \cap O)$$ $$P(W \cup O) = \frac{1}{2} + \frac{3}{8} - \frac{3}{20} = \frac{20}{40} + \frac{15}{40} - \frac{6}{40} = \frac{20 + 15 - 6}{40} = \frac{29}{40}$$
Final Answer: 29/40
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