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To determine which two events are independent, we need to check if the probabilities of their intersections equal the product of their individual probabilities. An event [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] are independent if and only if [tex]\( P(X \cap Y) = P(X) \cdot P(Y) \)[/tex].
Let's find the given probabilities step by step.
1. Calculate the total probabilities for each event:
- Event [tex]\( A \)[/tex] (Male): [tex]\( P(A) = \frac{36}{60} = 0.6 \)[/tex]
- Event [tex]\( B \)[/tex] (Female): [tex]\( P(B) = \frac{24}{60} = 0.4 \)[/tex]
- Event [tex]\( C \)[/tex] (Public Transportation): [tex]\( P(C) = \frac{20}{60} \approx 0.3333 \)[/tex]
- Event [tex]\( D \)[/tex] (Own Transportation): [tex]\( P(D) = \frac{30}{60} = 0.5 \)[/tex]
- Event [tex]\( E \)[/tex] (Other Transportation): [tex]\( P(E) = \frac{10}{60} \approx 0.1667 \)[/tex]
2. Calculate the intersection probabilities:
- Event [tex]\( A \cap C \)[/tex] (Male and Public Transportation): [tex]\( P(A \cap C) = \frac{12}{60} = 0.2 \)[/tex]
- Event [tex]\( A \cap D \)[/tex] (Male and Own Transportation): [tex]\( P(A \cap D) = \frac{20}{60} \approx 0.3333 \)[/tex]
- Event [tex]\( B \cap D \)[/tex] (Female and Own Transportation): [tex]\( P(B \cap D) = \frac{10}{60} \approx 0.1667 \)[/tex]
- Event [tex]\( B \cap E \)[/tex] (Female and Other Transportation): [tex]\( P(B \cap E) = \frac{6}{60} = 0.1 \)[/tex]
3. Check the independence conditions:
- For [tex]\( A \)[/tex] and [tex]\( C \)[/tex]:
[tex]\[ P(A \cap C) = 0.2 \qquad P(A) \cdot P(C) = 0.6 \cdot 0.3333 \approx 0.2 \][/tex]
[tex]\( \text{Since } P(A \cap C) \neq P(A) \cdot P(C), \text{ they are not independent.} \)[/tex]
- For [tex]\( A \)[/tex] and [tex]\( D \)[/tex]:
[tex]\[ P(A \cap D) \approx 0.3333 \qquad P(A) \cdot P(D) = 0.6 \cdot 0.5 = 0.3 \][/tex]
[tex]\( \text{Since } P(A \cap D) \neq P(A) \cdot P(D), \text{ they are not independent.} \)[/tex]
- For [tex]\( B \)[/tex] and [tex]\( D \)[/tex]:
[tex]\[ P(B \cap D) \approx 0.1667 \qquad P(B) \cdot P(D) = 0.4 \cdot 0.5 = 0.2 \][/tex]
[tex]\( \text{Since } P(B \cap D) \neq P(B) \cdot P(D), \text{ they are not independent.} \)[/tex]
- For [tex]\( B \)[/tex] and [tex]\( E \)[/tex]:
[tex]\[ P(B \cap E) = 0.1 \qquad P(B) \cdot P(E) = 0.4 \cdot 0.1667 \approx 0.0667 \][/tex]
[tex]\( \text{Since } P(B \cap E) \neq P(B) \cdot P(E), \text{ they are not independent.} \)[/tex]
Therefore, none of the given event pairs [tex]\((A \text{ and } C, A \text{ and } D, B \text{ and } D, B \text{ and } E)\)[/tex] are independent. Hence, there are no two independent events in the given data set.
Let's find the given probabilities step by step.
1. Calculate the total probabilities for each event:
- Event [tex]\( A \)[/tex] (Male): [tex]\( P(A) = \frac{36}{60} = 0.6 \)[/tex]
- Event [tex]\( B \)[/tex] (Female): [tex]\( P(B) = \frac{24}{60} = 0.4 \)[/tex]
- Event [tex]\( C \)[/tex] (Public Transportation): [tex]\( P(C) = \frac{20}{60} \approx 0.3333 \)[/tex]
- Event [tex]\( D \)[/tex] (Own Transportation): [tex]\( P(D) = \frac{30}{60} = 0.5 \)[/tex]
- Event [tex]\( E \)[/tex] (Other Transportation): [tex]\( P(E) = \frac{10}{60} \approx 0.1667 \)[/tex]
2. Calculate the intersection probabilities:
- Event [tex]\( A \cap C \)[/tex] (Male and Public Transportation): [tex]\( P(A \cap C) = \frac{12}{60} = 0.2 \)[/tex]
- Event [tex]\( A \cap D \)[/tex] (Male and Own Transportation): [tex]\( P(A \cap D) = \frac{20}{60} \approx 0.3333 \)[/tex]
- Event [tex]\( B \cap D \)[/tex] (Female and Own Transportation): [tex]\( P(B \cap D) = \frac{10}{60} \approx 0.1667 \)[/tex]
- Event [tex]\( B \cap E \)[/tex] (Female and Other Transportation): [tex]\( P(B \cap E) = \frac{6}{60} = 0.1 \)[/tex]
3. Check the independence conditions:
- For [tex]\( A \)[/tex] and [tex]\( C \)[/tex]:
[tex]\[ P(A \cap C) = 0.2 \qquad P(A) \cdot P(C) = 0.6 \cdot 0.3333 \approx 0.2 \][/tex]
[tex]\( \text{Since } P(A \cap C) \neq P(A) \cdot P(C), \text{ they are not independent.} \)[/tex]
- For [tex]\( A \)[/tex] and [tex]\( D \)[/tex]:
[tex]\[ P(A \cap D) \approx 0.3333 \qquad P(A) \cdot P(D) = 0.6 \cdot 0.5 = 0.3 \][/tex]
[tex]\( \text{Since } P(A \cap D) \neq P(A) \cdot P(D), \text{ they are not independent.} \)[/tex]
- For [tex]\( B \)[/tex] and [tex]\( D \)[/tex]:
[tex]\[ P(B \cap D) \approx 0.1667 \qquad P(B) \cdot P(D) = 0.4 \cdot 0.5 = 0.2 \][/tex]
[tex]\( \text{Since } P(B \cap D) \neq P(B) \cdot P(D), \text{ they are not independent.} \)[/tex]
- For [tex]\( B \)[/tex] and [tex]\( E \)[/tex]:
[tex]\[ P(B \cap E) = 0.1 \qquad P(B) \cdot P(E) = 0.4 \cdot 0.1667 \approx 0.0667 \][/tex]
[tex]\( \text{Since } P(B \cap E) \neq P(B) \cdot P(E), \text{ they are not independent.} \)[/tex]
Therefore, none of the given event pairs [tex]\((A \text{ and } C, A \text{ and } D, B \text{ and } D, B \text{ and } E)\)[/tex] are independent. Hence, there are no two independent events in the given data set.
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