The Bureau of Transportation Statistics reports on-time performance for airlines at major U.S. airports. JetBlue, United, and US Airways share Terminal C at Boston’s Logan Airport. The percentage of on-time flights reported for a sample month were 76.8% for JetBlue, 71.5% for United, and 82.2% for US Airways. Assume that 30% of the flights arriving at Terminal C are JetBlue flights, 32% are United flights, and 38% US Airways flights. a. Develop a joint probability table with three rows (the airlines) and two columns (on- time and late). b. An announcement is made that Flight 1382 will be arriving at gate 20 of Terminal C. What is the probability that Flight 1382 will arrive on time? c. What is the most likely airline for Flight 1382? What is the probability that Flight 1382 is by this airline? d. Suppose that an announcement is made saying that Flight 1382 will now be arriving late. What is the most likely airline for this flight? What is the probability that Flight 1382 is by this airline?
The Correct Answer and Explanation is:
Let’s solve each part of the question step by step:
a. Develop a joint probability table with three rows (the airlines) and two columns (on-time and late).
We are given the following information:
- JetBlue: 76.8% on-time, so the probability of a late flight for JetBlue is 1 – 0.768 = 0.232.
- United: 71.5% on-time, so the probability of a late flight for United is 1 – 0.715 = 0.285.
- US Airways: 82.2% on-time, so the probability of a late flight for US Airways is 1 – 0.822 = 0.178.
Also, we know the following probabilities for each airline:
- Probability of JetBlue flight: 0.30.
- Probability of United flight: 0.32.
- Probability of US Airways flight: 0.38.
Now, we can calculate the joint probabilities for each airline and whether the flight is on-time or late. The joint probability is the product of the airline’s probability and the flight’s on-time or late probability.
| Airline | On-Time (P(on-time)) | Late (P(late)) |
|---|---|---|
| JetBlue | 0.30 * 0.768 = 0.2304 | 0.30 * 0.232 = 0.0696 |
| United | 0.32 * 0.715 = 0.2288 | 0.32 * 0.285 = 0.0912 |
| US Airways | 0.38 * 0.822 = 0.3124 | 0.38 * 0.178 = 0.0676 |
The joint probability table is:
| Airline | On-Time | Late |
|---|---|---|
| JetBlue | 0.2304 | 0.0696 |
| United | 0.2288 | 0.0912 |
| US Airways | 0.3124 | 0.0676 |
b. Probability that Flight 1382 will arrive on time.
To find the probability that Flight 1382 will arrive on time, we sum up the on-time probabilities for all airlines:P(on-time)=P(JetBlue on-time)+P(United on-time)+P(US Airways on-time)P(\text{on-time}) = P(\text{JetBlue on-time}) + P(\text{United on-time}) + P(\text{US Airways on-time})P(on-time)=P(JetBlue on-time)+P(United on-time)+P(US Airways on-time)P(on-time)=0.2304+0.2288+0.3124=0.7716P(\text{on-time}) = 0.2304 + 0.2288 + 0.3124 = 0.7716P(on-time)=0.2304+0.2288+0.3124=0.7716
Thus, the probability that Flight 1382 will arrive on time is 0.7716 or 77.16%.
c. Most likely airline for Flight 1382.
To determine the most likely airline for Flight 1382, we need to calculate the posterior probabilities for each airline, given that the flight arrives on time. This is done using Bayes’ Theorem:P(JetBlue∣on-time)=P(JetBlue)⋅P(on-time∣JetBlue)P(on-time)P(\text{JetBlue} | \text{on-time}) = \frac{P(\text{JetBlue}) \cdot P(\text{on-time} | \text{JetBlue})}{P(\text{on-time})}P(JetBlue∣on-time)=P(on-time)P(JetBlue)⋅P(on-time∣JetBlue)P(United∣on-time)=P(United)⋅P(on-time∣United)P(on-time)P(\text{United} | \text{on-time}) = \frac{P(\text{United}) \cdot P(\text{on-time} | \text{United})}{P(\text{on-time})}P(United∣on-time)=P(on-time)P(United)⋅P(on-time∣United)P(US Airways∣on-time)=P(US Airways)⋅P(on-time∣US Airways)P(on-time)P(\text{US Airways} | \text{on-time}) = \frac{P(\text{US Airways}) \cdot P(\text{on-time} | \text{US Airways})}{P(\text{on-time})}P(US Airways∣on-time)=P(on-time)P(US Airways)⋅P(on-time∣US Airways)
Substituting the values:P(JetBlue∣on-time)=0.30⋅0.7680.7716=0.23040.7716≈0.299P(\text{JetBlue} | \text{on-time}) = \frac{0.30 \cdot 0.768}{0.7716} = \frac{0.2304}{0.7716} \approx 0.299P(JetBlue∣on-time)=0.77160.30⋅0.768=0.77160.2304≈0.299P(United∣on-time)=0.32⋅0.7150.7716=0.22880.7716≈0.297P(\text{United} | \text{on-time}) = \frac{0.32 \cdot 0.715}{0.7716} = \frac{0.2288}{0.7716} \approx 0.297P(United∣on-time)=0.77160.32⋅0.715=0.77160.2288≈0.297P(US Airways∣on-time)=0.38⋅0.8220.7716=0.31240.7716≈0.405P(\text{US Airways} | \text{on-time}) = \frac{0.38 \cdot 0.822}{0.7716} = \frac{0.3124}{0.7716} \approx 0.405P(US Airways∣on-time)=0.77160.38⋅0.822=0.77160.3124≈0.405
Thus, the most likely airline for Flight 1382 is US Airways, with a probability of 0.405 or 40.5%.
d. Most likely airline given that Flight 1382 is late.
Now, we need to calculate the posterior probabilities for each airline, given that the flight is late. We use Bayes’ Theorem again, but with the late probability.P(JetBlue∣late)=P(JetBlue)⋅P(late∣JetBlue)P(late)P(\text{JetBlue} | \text{late}) = \frac{P(\text{JetBlue}) \cdot P(\text{late} | \text{JetBlue})}{P(\text{late})}P(JetBlue∣late)=P(late)P(JetBlue)⋅P(late∣JetBlue)P(United∣late)=P(United)⋅P(late∣United)P(late)P(\text{United} | \text{late}) = \frac{P(\text{United}) \cdot P(\text{late} | \text{United})}{P(\text{late})}P(United∣late)=P(late)P(United)⋅P(late∣United)P(US Airways∣late)=P(US Airways)⋅P(late∣US Airways)P(late)P(\text{US Airways} | \text{late}) = \frac{P(\text{US Airways}) \cdot P(\text{late} | \text{US Airways})}{P(\text{late})}P(US Airways∣late)=P(late)P(US Airways)⋅P(late∣US Airways)
First, calculate the probability of being late:P(late)=P(JetBlue late)+P(United late)+P(US Airways late)P(\text{late}) = P(\text{JetBlue late}) + P(\text{United late}) + P(\text{US Airways late})P(late)=P(JetBlue late)+P(United late)+P(US Airways late)P(late)=0.0696+0.0912+0.0676=0.2284P(\text{late}) = 0.0696 + 0.0912 + 0.0676 = 0.2284P(late)=0.0696+0.0912+0.0676=0.2284
Now calculate the posterior probabilities for each airline:P(JetBlue∣late)=0.30⋅0.2320.2284=0.06960.2284≈0.305P(\text{JetBlue} | \text{late}) = \frac{0.30 \cdot 0.232}{0.2284} = \frac{0.0696}{0.2284} \approx 0.305P(JetBlue∣late)=0.22840.30⋅0.232=0.22840.0696≈0.305P(United∣late)=0.32⋅0.2850.2284=0.09120.2284≈0.400P(\text{United} | \text{late}) = \frac{0.32 \cdot 0.285}{0.2284} = \frac{0.0912}{0.2284} \approx 0.400P(United∣late)=0.22840.32⋅0.285=0.22840.0912≈0.400P(US Airways∣late)=0.38⋅0.1780.2284=0.06760.2284≈0.296P(\text{US Airways} | \text{late}) = \frac{0.38 \cdot 0.178}{0.2284} = \frac{0.0676}{0.2284} \approx 0.296P(US Airways∣late)=0.22840.38⋅0.178=0.22840.0676≈0.296
Thus, the most likely airline for Flight 1382, given that it is late, is United, with a probability of 0.400 or 40%.
Summary of answers:
- b. The probability that Flight 1382 will arrive on time is 77.16%.
- c. The most likely airline for Flight 1382 is US Airways, with a probability of 40.5%.
- d. The most likely airline for Flight 1382, given that it is late, is United, with a probability of 40%.
