Question

Three airlines serve a small town in Ohio. Airline A has 48% of all scheduled flights, airline B has 26% and airline C has the remaining 26%. Their on-time rates are 81%, 61%, and 40%, respectively. A flight just left on-time. What is the probability that it was a flight of airline A?

Answer 1

Answer:

0.5969 = 59.69% probability that it was a flight of airline A

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Left on time.

Event B: From airline A.

Probability of a flight leaving on time:

81% of 48%(airline A).

61% of 26%(airline B)

40% of 26%(airline C). So

$P(A) = 0.81*0.48 + 0.61*0.26 + 0.40*0.26 = 0.6514$

Probability of leaving on time and being from airline A:

81% of 48%. So

$P(A \cap B) = 0.81*0.48 = 0.3888$

What is the probability that it was a flight of airline A?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3888}{0.6514} = 0.5969$

0.5969 = 59.69% probability that it was a flight of airline A