Respuesta :
Answer:
46.67% probability that they flew with airline B.
Step-by-step explanation:
We have these following probabilities:
A 40% probability that a traveler chooses airline A.
A 35% probability that a traveled chooses airline B.
A 25% probability that a traveler chooses airline C.
If a passenger chooses airline A, a 10% probability that he arrives late.
If a passenger chooses airline B, a 15% probability that he arrives late.
If a passenger chooses airline C, a 8% probability that he arrives late.
If a randomly selected traveler is on a flight from Seattle which arrives late to Sydney, what is the probability that they flew with airline B?
This can be formulated as the following problem:
What is the probability of B happening, knowing that A has happened.
It can be calculated by the following formula
[tex]P = \frac{P(B).P(A/B)}{P(A)}[/tex]
Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.
So
What is the probability that the traveler flew with airline B, given that he was late?
P(B) is the probability that he flew with airline B.
So [tex]P(B) = 0.35[/tex]
P(A/B) is the probability of being late when traveling with airline B. So P(A/B) = 0.15.
P(A) is the probability of being late. This is the sum of 10% of 40%(airline A), 15% of 35%(airline B) and 8% of 25%(airline C).
So
[tex]P(A) = 0.1*0.4 + 0.15*0.35 + 0.08*0.25 = 0.1125[/tex]
Probability
[tex]P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.35*0.15}{0.1125} = 0.4667[/tex]
46.67% probability that they flew with airline B.
