Answer :
To determine whether two events are independent, we need to check if the probability of the intersection of the events is equal to the product of their individual probabilities. In mathematical terms, two events [tex]\(X\)[/tex] and [tex]\(Y\)[/tex] are independent if:
[tex]\[ P(X \cap Y) = P(X) \times P(Y) \][/tex]
Let's use the data provided in the table to test the independence of the given events. The table is as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|} \cline { 2 - 5 } \multicolumn{1}{c|}{} & \text{Public} & \text{Own} & \text{Other} & \text{Total} \\ \hline \text{Male} & 12 & 20 & 4 & 36 \\ \hline \text{Female} & 8 & 10 & 6 & 24 \\ \hline \text{Total} & 20 & 30 & 10 & 60 \\ \hline \end{array} \][/tex]
First, we need to convert these counts into probabilities.
### Event Probabilities
1. [tex]\( P(A) \)[/tex]: The probability that an employee is male.
[tex]\[ P(A) = \frac{36}{60} = 0.6 \][/tex]
2. [tex]\( P(B) \)[/tex]: The probability that an employee is female.
[tex]\[ P(B) = \frac{24}{60} = 0.4 \][/tex]
3. [tex]\( P(C) \)[/tex]: The probability that an employee takes public transportation.
[tex]\[ P(C) = \frac{20}{60} = 0.333 \][/tex]
4. [tex]\( P(D) \)[/tex]: The probability that an employee takes their own transportation.
[tex]\[ P(D) = \frac{30}{60} = 0.5 \][/tex]
5. [tex]\( P(E) \)[/tex]: The probability that an employee takes some other method of transportation.
[tex]\[ P(E) = \frac{10}{60} = 0.167 \][/tex]
### Intersections and Testing Independence
#### [tex]\( A \)[/tex] and [tex]\( C \)[/tex]:
1. Probability of being male and taking public transportation:
[tex]\[ P(A \cap C) = \frac{12}{60} = 0.2 \][/tex]
2. Check for independence:
[tex]\[ P(A) \times P(C) = 0.6 \times 0.333 = 0.2 \][/tex]
Result: [tex]\( A \)[/tex] and [tex]\( C \)[/tex] are not independent because [tex]\( 0.2 \ne 0.1998 \)[/tex].
#### [tex]\( A \)[/tex] and [tex]\( D \)[/tex]:
1. Probability of being male and taking own transportation:
[tex]\[ P(A \cap D) = \frac{20}{60} = 0.333 \][/tex]
2. Check for independence:
[tex]\[ P(A) \times P(D) = 0.6 \times 0.5 = 0.3 \][/tex]
Result: [tex]\( A \)[/tex] and [tex]\( D \)[/tex] are not independent because [tex]\( 0.333 \ne 0.3 \)[/tex].
#### [tex]\( B \)[/tex] and [tex]\( D \)[/tex]:
1. Probability of being female and taking own transportation:
[tex]\[ P(B \cap D) = \frac{10}{60} = 0.167 \][/tex]
2. Check for independence:
[tex]\[ P(B) \times P(D) = 0.4 \times 0.5 = 0.2 \][/tex]
Result: [tex]\( B \)[/tex] and [tex]\( D \)[/tex] are not independent because [tex]\( 0.167 \ne 0.2 \)[/tex].
#### [tex]\( B \)[/tex] and [tex]\( E \)[/tex]:
1. Probability of being female and taking other transportation:
[tex]\[ P(B \cap E) = \frac{6}{60} = 0.1 \][/tex]
2. Check for independence:
[tex]\[ P(B) \times P(E) = 0.4 \times 0.167 = 0.067 \][/tex]
Result: [tex]\( B \)[/tex] and [tex]\( E \)[/tex] are not independent because [tex]\( 0.1 \ne 0.067 \)[/tex].
### Conclusion
None of the event pairs [tex]\( (A, C) \)[/tex], [tex]\( (A, D) \)[/tex], [tex]\( (B, D) \)[/tex], and [tex]\( (B, E) \)[/tex] are independent.
[tex]\[ P(X \cap Y) = P(X) \times P(Y) \][/tex]
Let's use the data provided in the table to test the independence of the given events. The table is as follows:
[tex]\[ \begin{array}{|c|c|c|c|c|} \cline { 2 - 5 } \multicolumn{1}{c|}{} & \text{Public} & \text{Own} & \text{Other} & \text{Total} \\ \hline \text{Male} & 12 & 20 & 4 & 36 \\ \hline \text{Female} & 8 & 10 & 6 & 24 \\ \hline \text{Total} & 20 & 30 & 10 & 60 \\ \hline \end{array} \][/tex]
First, we need to convert these counts into probabilities.
### Event Probabilities
1. [tex]\( P(A) \)[/tex]: The probability that an employee is male.
[tex]\[ P(A) = \frac{36}{60} = 0.6 \][/tex]
2. [tex]\( P(B) \)[/tex]: The probability that an employee is female.
[tex]\[ P(B) = \frac{24}{60} = 0.4 \][/tex]
3. [tex]\( P(C) \)[/tex]: The probability that an employee takes public transportation.
[tex]\[ P(C) = \frac{20}{60} = 0.333 \][/tex]
4. [tex]\( P(D) \)[/tex]: The probability that an employee takes their own transportation.
[tex]\[ P(D) = \frac{30}{60} = 0.5 \][/tex]
5. [tex]\( P(E) \)[/tex]: The probability that an employee takes some other method of transportation.
[tex]\[ P(E) = \frac{10}{60} = 0.167 \][/tex]
### Intersections and Testing Independence
#### [tex]\( A \)[/tex] and [tex]\( C \)[/tex]:
1. Probability of being male and taking public transportation:
[tex]\[ P(A \cap C) = \frac{12}{60} = 0.2 \][/tex]
2. Check for independence:
[tex]\[ P(A) \times P(C) = 0.6 \times 0.333 = 0.2 \][/tex]
Result: [tex]\( A \)[/tex] and [tex]\( C \)[/tex] are not independent because [tex]\( 0.2 \ne 0.1998 \)[/tex].
#### [tex]\( A \)[/tex] and [tex]\( D \)[/tex]:
1. Probability of being male and taking own transportation:
[tex]\[ P(A \cap D) = \frac{20}{60} = 0.333 \][/tex]
2. Check for independence:
[tex]\[ P(A) \times P(D) = 0.6 \times 0.5 = 0.3 \][/tex]
Result: [tex]\( A \)[/tex] and [tex]\( D \)[/tex] are not independent because [tex]\( 0.333 \ne 0.3 \)[/tex].
#### [tex]\( B \)[/tex] and [tex]\( D \)[/tex]:
1. Probability of being female and taking own transportation:
[tex]\[ P(B \cap D) = \frac{10}{60} = 0.167 \][/tex]
2. Check for independence:
[tex]\[ P(B) \times P(D) = 0.4 \times 0.5 = 0.2 \][/tex]
Result: [tex]\( B \)[/tex] and [tex]\( D \)[/tex] are not independent because [tex]\( 0.167 \ne 0.2 \)[/tex].
#### [tex]\( B \)[/tex] and [tex]\( E \)[/tex]:
1. Probability of being female and taking other transportation:
[tex]\[ P(B \cap E) = \frac{6}{60} = 0.1 \][/tex]
2. Check for independence:
[tex]\[ P(B) \times P(E) = 0.4 \times 0.167 = 0.067 \][/tex]
Result: [tex]\( B \)[/tex] and [tex]\( E \)[/tex] are not independent because [tex]\( 0.1 \ne 0.067 \)[/tex].
### Conclusion
None of the event pairs [tex]\( (A, C) \)[/tex], [tex]\( (A, D) \)[/tex], [tex]\( (B, D) \)[/tex], and [tex]\( (B, E) \)[/tex] are independent.
