Brown Law Firm collected data on the transportation choices of its employees for their morning commute. The table shows the percentages of the type of transportation of the male and female employees.

[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline
& Public & Own & Other & Total \\
\hline
Male & 12 & 20 & 4 & 36 \\
\hline
Female & 8 & 10 & 6 & 24 \\
\hline
Total & 20 & 30 & 10 & 60 \\
\hline
\end{tabular}
\][/tex]

Consider the following events:
- A: The employee is male.
- B: The employee is female.
- C: The employee takes public transportation.
- D: The employee takes his/her own transportation.
- E: The employee takes some other method of transportation.

Which two events are independent?
A. A and C
B. A and D
C. B and D
D. B and E



Answer :

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.