Answer to Question #230460 in Statistics and Probability for Ankur Sinha

Question

Three airlines serve a small town in Rajkot District. Airline A has 50% of all the scheduled

flights, airline B has 30%, and airline C has the remaining 20%. Their on-time rates are 80%, 65%,

and 40%, respectively.

A plane has just left on time. What is the probability that it was airline A?

Note: Draw the Tree-diagram using Paint App.

Answer 1

Solution:

Let us consider,

The event A represents that scheduled flights for Airline A.

The event B represents that scheduled flights for Airline B.

The event C represents that scheduled flights for Airline C.

The event E represents that flight left in on-time.

The prior probabilities are

P(A)=0.50

P(B)=0.30

P(C)=0.20

The posterior probabilities are

P(E|A)=0.80

P(E|B)=0.65

P(E|C)=0.40

We have to find the probability that it was airline A, if the plane has just left on time, i.e., P(A|E)

Substituting the prior and likelihood (posterior) probabilities into the Bayes’s formula, then it yields

$P(A|E) = \frac{P(E|A) \times P(A)}{P(E|A) \times P(A) + P(E|B) \times P(B) + P(E|C) \times P(C)} \\ = \frac{0.80 \times 0.50}{0.80 \times 0.50 + 0.65 \times 0.30 + 0.40 \times 0.20} \\ = \frac{0.4}{0.675} \\ = 0.59259$

Therefore, the probability that it was airline A, if the plane has just left on time is 0.593.