Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
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.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
