To determine which event has the highest probability, we will calculate the probabilities of each event.
First, we need to understand the total number of buses observed, which is 110.
Event A: The bus is from Route C and is on time.
- Route C on-time buses: 24
- Total buses: 110
[tex]\[ \text{Probability} = \frac{\text{Route C on-time}}{\text{Total buses}} = \frac{24}{110} \approx 0.218 \][/tex]
Event B: The bus is from Route B and is delayed.
- Route B delayed buses: 8
- Total buses: 110
[tex]\[ \text{Probability} = \frac{\text{Route B delayed}}{\text{Total buses}} = \frac{8}{110} \approx 0.073 \][/tex]
Event C: The bus is from Route C and is delayed.
- Route C delayed buses: 6
- Total buses: 110
[tex]\[ \text{Probability} = \frac{\text{Route C delayed}}{\text{Total buses}} = \frac{6}{110} \approx 0.055 \][/tex]
Event D: The bus is from Route A and is on time.
- Route A on-time buses: 28
- Total buses: 110
[tex]\[ \text{Probability} = \frac{\text{Route A on-time}}{\text{Total buses}} = \frac{28}{110} \approx 0.255 \][/tex]
Comparing the probabilities:
- Probability of Event A (Route C on time): [tex]\( \approx 0.218 \)[/tex]
- Probability of Event B (Route B delayed): [tex]\( \approx 0.073 \)[/tex]
- Probability of Event C (Route C delayed): [tex]\( \approx 0.055 \)[/tex]
- Probability of Event D (Route A on time): [tex]\( \approx 0.255 \)[/tex]
The highest probability is [tex]\( \approx 0.255 \)[/tex], which corresponds to Event D: The bus is from Route A and is on time.
Therefore, the correct answer is:
D. The bus is from route [tex]$A$[/tex] and is on time.
