Respuesta :
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
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] 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
[tex]P(A) = 0.81*0.48 + 0.61*0.26 + 0.40*0.26 = 0.6514[/tex]
Probability of leaving on time and being from airline A:
81% of 48%. So
[tex]P(A \cap B) = 0.81*0.48 = 0.3888[/tex]
What is the probability that it was a flight of airline A?
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.3888}{0.6514} = 0.5969[/tex]
0.5969 = 59.69% probability that it was a flight of airline A
