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]
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]
