Question

select the correct answer.

for a one - week period, three bus routes were observed. the results are shown in the table below.

	on - time	delayed	total
bus route b	37	8	45
bus route c	24	6	30
total	89	21	110

a bus is selected randomly. which event has the highest probability?

a. the bus is from route b and is delayed.

b. the bus is from route a and is on time.

c. the bus is from route c and is on time.

d. the bus is from route c and is delayed.

Explanation:

Step1: Calculate Probability for A

Probability = $\frac{\text{Route B delayed}}{\text{Total buses}} = \frac{8}{110} \approx 0.0727$

Step2: Calculate Probability for B

Probability = $\frac{\text{Route A on-time}}{\text{Total buses}} = \frac{28}{110} \approx 0.2545$

Step3: Calculate Probability for C

Probability = $\frac{\text{Route C on-time}}{\text{Total buses}} = \frac{24}{110} \approx 0.2182$

Step4: Calculate Probability for D

Probability = $\frac{\text{Route C delayed}}{\text{Total buses}} = \frac{6}{110} \approx 0.0545$

Step5: Compare all probabilities

$0.2545 > 0.2182 > 0.0727 > 0.0545$

Answer:

B. The bus is from route A and is on time.