Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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]
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
