Me and my group have been struggling to figure out this prob

Me and my group have been struggling to figure out this probability problem. So please show us a step by step solution to it and if possible, explain the logic as it is equally as important as the answer.

There are four major air carriers with flights from Boston to Los Angeles, 32% of all passengers take American Airlines, 25% take Jet Blue, 17% take United, and 26% take Virgin America. Data from 2012 indicate that 20% of all American Airlines flights from Boston to Lost Angeles are late, 23% of Jet Blue frights from Boston to Los Angeles are late, 19% of United flights from Boston to Los Angeles are late, and 1 1% of Virgin America flights from Boston to Los Angeles are late. Suppose a passenger taking a flight from Boston to Los Angeles is randomly selected. a. What is the probability that the passenger takes American Airlines and is late? b. What is the probability that the passenger is on time in general? Which airline did the passenger most likely fly? Show proof for your Suppose the passenger arrives late. answer (ie. Discuss all of the relevant probabilities) c.

Solution

a) Prob that the passenger takes American air lines and is late =

Given that 32% take American airlines.and 20% of American airlines are late.

Hence as events are independent,

reqd prob = 32%(20%) = 640/10000 = 0.064

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b) prob for passenger on time in general

This is possible if he takes any one flight and that flight arrives in time.

i.e. 32%(100-20)% + 25%(100-23)%+17%(100-19)%+26%(100-11)%

= 0.256+0.1925+0.1377+0.2314

= 0.8176

c) Given that a passenger arrives late.

Let us find prob of late for all 4 planes.

i) American airlines Prob of late = 0.32(0.20) = 0.064

ii) Jet Blue Prob for late             = 0.25(0.23) =0.0575

iii) United Prob for late               = 0.17(0.19) = 0.0323

iv) Virgin Americal Pr for late      = 0.26(0.11) = 0.0286

COmparing the four, we find that Jet blue has more prob.

Hence more likely he must have selected Jet blue.