Question

select the correct answer.
for a one - week period, three bus routes were observed. the results are shown in the table below
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 c and is on time.
c. the bus is from route a and is on time.
d. the bus is from route c and is delayed.

Explanation:

Step1: Calculate probability for each option

Probability formula: \(P=\frac{\text{Favorable outcomes}}{\text{Total outcomes}}\)

• Option A: Bus is from route B and is delayed. Favorable outcomes \( = 8\), Total outcomes \(=110\). \(P_A=\frac{8}{110}\approx0.0727\)
• Option B: Bus is from route C and is on - time. Favorable outcomes \( = 24\), Total outcomes \(=110\). \(P_B=\frac{24}{110}\approx0.2182\)
• Option C: Bus is from route A and is on - time. Favorable outcomes \( = 28\), Total outcomes \(=110\). \(P_C=\frac{28}{110}\approx0.2545\)
• Option D: Bus is from route C and is delayed. Favorable outcomes \( = 6\), Total outcomes \(=110\). \(P_D=\frac{6}{110}\approx0.0545\)

Step2: Compare probabilities

\(P_D\approx0.0545\lt P_A\approx0.0727\lt P_B\approx0.2182\lt P_C\approx0.2545\)

Answer:

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