Identifying Two Independent Events

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.

\begin{tabular}{|c|c|c|c|c|}
\cline { 2 - 4 } \multicolumn{1}{c|}{} & 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}

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

Which two events are independent?

A. [tex]\( A \)[/tex] and [tex]\( C \)[/tex]

B. [tex]\( A \)[/tex] and [tex]\( D \)[/tex]

C. [tex]\( B \)[/tex] and [tex]\( D \)[/tex]

D. [tex]\( B \)[/tex] and [tex]\( E \)[/tex]



Answer :

To determine the independence of events, we can use the concept of conditional probability. Two events [tex]\( X \)[/tex] and [tex]\( Y \)[/tex] are independent if and only if [tex]\( P(X \cap Y) = P(X) \cdot P(Y) \)[/tex], or equivalently, [tex]\( P(X \mid Y) = P(X) \)[/tex].

Let’s examine the provided data:

### Probabilities of Individual Events
1. Total number of employees: 60
2. Probability of an employee being male ([tex]\( P(A) \)[/tex]):
[tex]\[ P(A) = \frac{36}{60} = 0.6 \][/tex]
3. Probability of an employee being female ([tex]\( P(B) \)[/tex]):
[tex]\[ P(B) = \frac{24}{60} = 0.4 \][/tex]
4. Probability of an employee taking public transportation ([tex]\( P(C) \)[/tex]):
[tex]\[ P(C) = \frac{20}{60} = 0.3333 \][/tex]
5. Probability of an employee taking their own transportation ([tex]\( P(D) \)[/tex]):
[tex]\[ P(D) = \frac{30}{60} = 0.5 \][/tex]
6. Probability of an employee taking other forms of transportation ([tex]\( P(E) \)[/tex]):
[tex]\[ P(E) = \frac{10}{60} = 0.1667 \][/tex]

### Conditional Probabilities
1. Probability that an employee is male given that they take public transportation ([tex]\( P(A \mid C) \)[/tex]):
- 12 out of 20 public transport users are male:
[tex]\[ P(A \mid C) = \frac{12}{20} = 0.6 \][/tex]
Compare this with [tex]\( P(A) \)[/tex]:
[tex]\[ P(A \mid C) = P(A) \quad (\text{True}) \][/tex]
Thus, events [tex]\( A \)[/tex] and [tex]\( C \)[/tex] are independent.

2. Probability that an employee is male given that they take their own transportation ([tex]\( P(A \mid D) \)[/tex]):
- 20 out of 30 own transport users are male:
[tex]\[ P(A \mid D) = \frac{20}{30} = 0.6667 \][/tex]
Compare this with [tex]\( P(A) \)[/tex]:
[tex]\[ P(A \mid D) \neq P(A) \quad (\text{False}) \][/tex]
Thus, events [tex]\( A \)[/tex] and [tex]\( D \)[/tex] are not independent.

3. Probability that an employee is female given that they take their own transportation ([tex]\( P(B \mid D) \)[/tex]):
- 10 out of 30 own transport users are female:
[tex]\[ P(B \mid D) = \frac{10}{30} = 0.3333 \][/tex]
Compare this with [tex]\( P(B) \)[/tex]:
[tex]\[ P(B \mid D) \neq P(B) \quad (\text{False}) \][/tex]
Thus, events [tex]\( B \)[/tex] and [tex]\( D \)[/tex] are not independent.

4. Probability that an employee is female given that they use other forms of transport ([tex]\( P(B \mid E) \)[/tex]):
- 6 out of 10 other transport users are female:
[tex]\[ P(B \mid E) = \frac{6}{10} = 0.6 \][/tex]
Compare this with [tex]\( P(B) \)[/tex]:
[tex]\[ P(B \mid E) \neq P(B) \quad (\text{False}) \][/tex]
Thus, events [tex]\( B \)[/tex] and [tex]\( E \)[/tex] are not independent.

Therefore, the only pair of independent events from the choices given is:

[tex]\[ A \text{ and } C \: (\text{The employee is male and takes public transportation}) \][/tex]

So, the two events that are independent are:
[tex]\[ \boxed{A \text{ and } C} \][/tex]