Question

A person traveling from Seattle to Sydney has three airlines to choose from. 40% of travelers choose airline A, and this airline arrives late to Sydney 10% of the time. 35% choose airline B, and this airline arrives late 15% of the time. 25% choose airline C, and this airline is late 8% of the time. 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?

Answer 1

46.67% probability that they flew with airline B.

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

`P = (P(B).P(A/B))/(P(A))`

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
`P(B) = 0.35`

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

`P(A) = 0.1*0.4 + 0.15*0.35 + 0.08*0.25 = 0.1125`

Probability

`P = (P(B).P(A/B))/(P(A)) = (0.35*0.15)/(0.1125) = 0.4667`

46.67% probability that they flew with airline B.