To determine if events A (late arrival) and B (transportation by car) are independent, we need to verify the condition:
\[ P(A \cap B) = P(A) \cdot P(B) \]
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Calculating the probabilities:
Let's find \( P(A) \) (the probability of a late arrival), \( P(B) \) (the probability of being transported by car), and \( P(A \cap B) \) (the probability of being late and using a car).
Given:
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Total number of people = \( 150 \) (late) + \( 250 - 20 \) (on time with a car) + \( 20 \) (late with a car) = \( 150 + 250 = 400 \).
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Number of people transported by car = \( 250 \).
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Number of people who were late with a car = \( 20 \).
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Calculating \( P(A) \):
\[ P(A) = \frac{\text{Number of late people}}{\text{Total number of people}} = \frac{150}{400} = 0.375 \]
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Calculating \( P(B) \):
\[ P(B) = \frac{\text{Number of people using a car}}{\text{Total number of people}} = \frac{250}{400} = 0.625 \]
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Calculating \( P(A \cap B) \):
\[ P(A \cap B) = \frac{\text{Number of late people using a car}}{\text{Total number of people}} = \frac{20}{400} = 0.05 \]
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Checking independence:
Now we need to check if \( P(A \cap B) = P(A) \cdot P(B) \):
\[ P(A) \cdot P(B) = 0.375 \cdot 0.625 = 0.234375 \]
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Conclusion:
Since:
\[ P(A \cap B) = 0.05 \quad \text{and} \quad P(A) \cdot P(B) = 0.234375 \]
These two values are not equal, hence events A and B are not independent.
Thus, for events A and B to be independent, the product \( P(A) \cdot P(B) \) would need to equal \( 0.05 \). Rounding to the nearest hundredth, that is:
\[ \boxed{0.05} \]