- Identify the probability of each event individually and their intersections from the table.
- Use the independence condition: two events \(X\) and \(Y\) are independent if \(P(X \cap Y) = P(X) \cdot P(Y)\).
- Evaluate the condition for each given pair:
- For A and C: \(P(A \cap C) = \frac{12}{60} = 0.2\) and \(P(A) \cdot P(C) = 0.6 \cdot \frac{1}{3} = 0.2\). Matches!
- The two independent events are \(\boxed{A \text{ and } C}\).
Explanation
- Analyze the problem and calculate individual event probabilities
To determine if two events \(X\) and \(Y\) are independent, we use the definition that states the events are independent if and only if the probability of both occurring together, \(P(X \cap Y)\), is equal to the product of their individual probabilities, \(P(X) \cdot P(Y)\). We are given a total of 60 employees. We identify the individual event probabilities first:
- \(P(A) = \frac{36}{60} = 0.6\)
- \(P(B) = \frac{24}{60} = 0.4\)
- \(P(C) = \frac{20}{60} = \frac{1}{3} \approx 0.333\)
- \(P(D) = \frac{30}{60} = 0.5\)
- \(P(E) = \frac{10}{60} = \frac{1}{6} \approx 0.167\)
- Test each pair of events for independence
We now test each pair provided in the options by comparing \(P(X \cap Y)\) to \(P(X) \cdot P(Y)\):
- For A and C:
- \(P(A \cap C) = \frac{12}{60} = 0.2\)
- \(P(A) \cdot P(C) = 0.6 \cdot \frac{1}{3} = 0.2\)
Since \(P(A \cap C) = P(A) \cdot P(C)\), these events are independent.
- For A and D:
- \(P(A \cap D) = \frac{20}{60} = \frac{1}{3} \approx 0.333\)
- \(P(A) \cdot P(D) = 0.6 \cdot 0.5 = 0.3\)
Since \(0.333
eq 0.3\), these events are not independent.
- For B and D:
- \(P(B \cap D) = \frac{10}{60} = \frac{1}{6} \approx 0.167\)
- \(P(B) \cdot P(D) = 0.4 \cdot 0.5 = 0.2\)
Since \(0.167
eq 0.2\), these events are not independent.
- For B and E:
- \(P(B \cap E) = \frac{6}{60} = 0.1\)
- \(P(B) \cdot P(E) = 0.4 \cdot \frac{1}{6} \approx 0.067\)
Since \(0.1
eq 0.067\), these events are not independent.
- Conclusion
Based on the calculations above, the only pair of events that satisfies the mathematical definition of independence is A and C. Therefore, we conclude that being male and taking public transportation are independent events in this data set.
Examples
Understanding whether two events are independent is crucial in data analysis, such as determining if a specific employee demographic prefers a particular transportation method. For instance, a firm might use this to analyze if commute choices are influenced by gender, which can help in planning corporate benefits like parking facilities or transit subsidies. Suppose a company has 100 employees, and 40 are male, while 20 take public transit. If 8 are both male and take public transit, we check for independence by testing if \(P(M \cap PT) = P(M) \cdot P(PT)\). Here, \(P(M) = \frac{40}{100} = 0.4\), \(P(PT) = \frac{20}{100} = 0.2\), and \(P(M \cap PT) = \frac{8}{100} = 0.08\). Since \(0.4 \times 0.2 = 0.08\), the events are indeed independent, meaning gender does not statistically correlate with public transit use in this hypothetical company.