The Bureau of Transportation Statistics reports On-time performance for airlines at ma- jor airports_ JetBlue United, and US Airways share terminal € at Boston’s Logan Airport. Suppose that the Percentage of on-time flights reported was 76.8% for JetBlue; 71.5% for United and 82.27 for US Airways_ Assume that 307 of flights arriving at terminal arc JetBlue flights; 32% are United Flights, and 38% are US Airways Flights. a) Develop joint probability table with threc rows (the airlines) and two columns (on timc status)_ Round the the fourth decimal place b) An announcement is made that Flight 1382 will 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? Suppose that 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 break down the problem step by step:
a) Joint Probability Table
We are given:
- JetBlue: 76.8% on-time, 307 flights at the terminal.
- United: 71.5% on-time, 32% of flights at the terminal.
- US Airways: 82.2% on-time, 38% of flights at the terminal.
From the data, we can construct the joint probability table.
Total percentage of flights:
- The total percentage of flights at the terminal is 307 for JetBlue, 32% for United, and 38% for US Airways. This gives us the proportions of flights at each airline’s terminal.
- P(JetBlue) = 0.307
- P(United) = 0.32
- P(US Airways) = 0.38
Now, for each airline, we need to calculate the probability of on-time and late arrivals.
JetBlue:
- P(On-time | JetBlue) = 0.768
- P(Late | JetBlue) = 1 – 0.768 = 0.232
United:
- P(On-time | United) = 0.715
- P(Late | United) = 1 – 0.715 = 0.285
US Airways:
- P(On-time | US Airways) = 0.822
- P(Late | US Airways) = 1 – 0.822 = 0.178
We can now calculate the joint probabilities by multiplying the marginal probabilities by the conditional probabilities (on-time or late):P(JetBlue and On-time)=P(JetBlue)×P(On-time | JetBlue)=0.307×0.768=0.2362\text{P(JetBlue and On-time)} = P(\text{JetBlue}) \times P(\text{On-time | JetBlue}) = 0.307 \times 0.768 = 0.2362P(JetBlue and On-time)=P(JetBlue)×P(On-time | JetBlue)=0.307×0.768=0.2362P(JetBlue and Late)=P(JetBlue)×P(Late | JetBlue)=0.307×0.232=0.0712\text{P(JetBlue and Late)} = P(\text{JetBlue}) \times P(\text{Late | JetBlue}) = 0.307 \times 0.232 = 0.0712P(JetBlue and Late)=P(JetBlue)×P(Late | JetBlue)=0.307×0.232=0.0712P(United and On-time)=P(United)×P(On-time | United)=0.32×0.715=0.2288\text{P(United and On-time)} = P(\text{United}) \times P(\text{On-time | United}) = 0.32 \times 0.715 = 0.2288P(United and On-time)=P(United)×P(On-time | United)=0.32×0.715=0.2288P(United and Late)=P(United)×P(Late | United)=0.32×0.285=0.0912\text{P(United and Late)} = P(\text{United}) \times P(\text{Late | United}) = 0.32 \times 0.285 = 0.0912P(United and Late)=P(United)×P(Late | United)=0.32×0.285=0.0912P(US Airways and On-time)=P(US Airways)×P(On-time | US Airways)=0.38×0.822=0.3124\text{P(US Airways and On-time)} = P(\text{US Airways}) \times P(\text{On-time | US Airways}) = 0.38 \times 0.822 = 0.3124P(US Airways and On-time)=P(US Airways)×P(On-time | US Airways)=0.38×0.822=0.3124P(US Airways and Late)=P(US Airways)×P(Late | US Airways)=0.38×0.178=0.0676\text{P(US Airways and Late)} = P(\text{US Airways}) \times P(\text{Late | US Airways}) = 0.38 \times 0.178 = 0.0676P(US Airways and Late)=P(US Airways)×P(Late | US Airways)=0.38×0.178=0.0676
Joint Probability Table:
| Airline | On-time (P) | Late (P) |
|---|---|---|
| JetBlue | 0.2362 | 0.0712 |
| United | 0.2288 | 0.0912 |
| US Airways | 0.3124 | 0.0676 |
b) Probability that Flight 1382 is on time
To find the probability that Flight 1382 will arrive on time, we need to use the law of total probability:P(On-time)=P(On-time | JetBlue)×P(JetBlue)+P(On-time | United)×P(United)+P(On-time | US Airways)×P(US Airways)P(\text{On-time}) = P(\text{On-time | JetBlue}) \times P(\text{JetBlue}) + P(\text{On-time | United}) \times P(\text{United}) + P(\text{On-time | US Airways}) \times P(\text{US Airways})P(On-time)=P(On-time | JetBlue)×P(JetBlue)+P(On-time | United)×P(United)+P(On-time | US Airways)×P(US Airways)
Substitute the values:P(On-time)=(0.768×0.307)+(0.715×0.32)+(0.822×0.38)P(\text{On-time}) = (0.768 \times 0.307) + (0.715 \times 0.32) + (0.822 \times 0.38)P(On-time)=(0.768×0.307)+(0.715×0.32)+(0.822×0.38)P(On-time)=0.2362+0.2288+0.3124=0.7774P(\text{On-time}) = 0.2362 + 0.2288 + 0.3124 = 0.7774P(On-time)=0.2362+0.2288+0.3124=0.7774
So, the probability that Flight 1382 will arrive on time is 0.7774 or 77.74%.
c) Most likely airline for Flight 1382 (if on time)
To determine the most likely airline for Flight 1382 if it arrives on time, we need to find the conditional probabilities using Bayes’ Theorem.
The conditional probability is given by:P(JetBlue | On-time)=P(JetBlue and On-time)P(On-time)=0.23620.7774=0.303P(\text{JetBlue | On-time}) = \frac{P(\text{JetBlue and On-time})}{P(\text{On-time})} = \frac{0.2362}{0.7774} = 0.303P(JetBlue | On-time)=P(On-time)P(JetBlue and On-time)=0.77740.2362=0.303P(United | On-time)=P(United and On-time)P(On-time)=0.22880.7774=0.294P(\text{United | On-time}) = \frac{P(\text{United and On-time})}{P(\text{On-time})} = \frac{0.2288}{0.7774} = 0.294P(United | On-time)=P(On-time)P(United and On-time)=0.77740.2288=0.294P(US Airways | On-time)=P(US Airways and On-time)P(On-time)=0.31240.7774=0.402P(\text{US Airways | On-time}) = \frac{P(\text{US Airways and On-time})}{P(\text{On-time})} = \frac{0.3124}{0.7774} = 0.402P(US Airways | On-time)=P(On-time)P(US Airways and On-time)=0.77740.3124=0.402
The most likely airline is US Airways, with a probability of 40.2%.
d) Most likely airline if Flight 1382 is late
For the late arrival scenario, we repeat the same steps but with the probabilities of late arrivals.
Conditional probabilities for late flights:
P(JetBlue | Late)=P(JetBlue and Late)P(Late)=0.07121−0.7774=0.07120.2226=0.32P(\text{JetBlue | Late}) = \frac{P(\text{JetBlue and Late})}{P(\text{Late})} = \frac{0.0712}{1 – 0.7774} = \frac{0.0712}{0.2226} = 0.32P(JetBlue | Late)=P(Late)P(JetBlue and Late)=1−0.77740.0712=0.22260.0712=0.32P(United | Late)=P(United and Late)P(Late)=0.09120.2226=0.409P(\text{United | Late}) = \frac{P(\text{United and Late})}{P(\text{Late})} = \frac{0.0912}{0.2226} = 0.409P(United | Late)=P(Late)P(United and Late)=0.22260.0912=0.409P(US Airways | Late)=P(US Airways and Late)P(Late)=0.06760.2226=0.303P(\text{US Airways | Late}) = \frac{P(\text{US Airways and Late})}{P(\text{Late})} = \frac{0.0676}{0.2226} = 0.303P(US Airways | Late)=P(Late)P(US Airways and Late)=0.22260.0676=0.303
The most likely airline if Flight 1382 is late is United, with a probability of 40.9%.
Summary:
- a) The joint probability table has been developed.
- b) The probability that Flight 1382 will arrive on time is 77.74%.
- c) The most likely airline for Flight 1382 to be on time is US Airways with a probability of 40.2%.
- d) If Flight 1382 is late, the most likely airline is United with a probability of 40.9%.
