Question

Bob can go to work by one of four modes of transportation: 1.) bicycle, 2.) car, 3.) bus and 4.) train. If he takes his bicycle, Bob has a 25% chance of getting to work on time. Because of heavy traffic, there is a 43% chance of being late if he takes his car. If he goes by bus, which has special reserve lanes to help keep schedule, but can be over crowded forcing him to wait for the next bus, he has a 15% chance of being late. The train runs on a very specific schedule, so there is only 5% probability of being late if he decides to take the train. Suppose that Bob is late and his boss see's Bob coming in late. His boss tries to estimate the probability Bob took his car in. Not knowing Bob's travel habbits, what does the boss estimate Bob's probability of taking the car given that he was late.

Answer 1

Answer:

the probability is 0.311 (31.1%)

Step-by-step explanation:

defining the event L= being late to work :Then knowing that each mode of transportation is equally likely (since we do not know its travel habits) :

P(L)= probability of taking the bicycle * probability of being late if he takes the bicycle + probability of taking the car* probability of being late if he takes the car  + probability of taking the bus* probability of being late if he takes the bus +probability of taking the train* probability of being late if he takes the train = 1/4  * 0.75 +   1/4 * 0.43 + 1/4  * 0.15  + 1/4 * 0.05 = 0.345

then we can use the theorem of Bayes for conditional probability. Thus defining the event C= Bob takes the car , we have

P(C/L)=  P(C∩L)/P(L) = 1/4 * 0.43 /0.345 = 0.311 (31.1%)

where

P(C∩L)= probability of taking the car and being late

P(C/L)= probability that Bob had taken the car given that he is late