To solve this problem, let's calculate the relevant probabilities based on the data provided in the table.
1. Probability of an employee receiving a speeding ticket given that they regularly leave for work early or on time:
- Number of employees who leave early or on time and received a speeding ticket = 5
- Total number of employees who leave early or on time = 86
The probability is calculated as follows:
[tex]\[
P(\text{Ticket | Early}) = \frac{\text{Number of employees with ticket who leave early}}{\text{Total number of employees who leave early}}
\][/tex]
[tex]\[
P(\text{Ticket | Early}) = \frac{5}{86} \approx 0.0581
\][/tex]
2. Probability of an employee not receiving a speeding ticket given that they regularly leave for work late:
- Number of employees who leave late and did not receive a speeding ticket = 9
- Total number of employees who leave late = 65
The probability is calculated as follows:
[tex]\[
P(\text{No Ticket | Late}) = \frac{\text{Number of employees without ticket who leave late}}{\text{Total number of employees who leave late}}
\][/tex]
[tex]\[
P(\text{No Ticket | Late}) = \frac{9}{65} \approx 0.1385
\][/tex]
3. Comparison of Probabilities:
- From our calculations, we have:
- [tex]\( P(\text{Ticket | Early}) \approx 0.0581 \)[/tex]
- [tex]\( P(\text{No Ticket | Late}) \approx 0.1385 \)[/tex]
Comparing these two probabilities, we find that:
[tex]\[
P(\text{Ticket | Early}) < P(\text{No Ticket | Late})
\][/tex]
Based on the comparison, we can make the following conclusion:
A. The probability of an employee receiving a speeding ticket given that they regularly leave for work early or on time is less than the probability of an employee not receiving a speeding ticket given that they regularly leave for work late.
So, the correct conclusion is Option A.
