Suppose that Bob can decide to go to work by one of three modes of transportatio

Question

Suppose that Bob can decide to go to work by one of three modes of transportation, car, bus, or commuter train. Because of high traffic, if he decides to go by car, there is a 50% chance he will be late. If he goes by bus, which has special reserved lanes but is sometimes overcrowded, the probability of being late is only 20%. The commuter train is almost never late, with a probability of only 1%, but is more expensive than the bus. Suppose that a coworker of Bob's knows that he almost always takes the commuter train to work, never takes the bus, but sometimes, 10% of the time, takes the car.

What is the coworkers probability that Bob drove to work that day, given that he was late?

Explanation / Answer

Let Event A = Bob Drove to work

Let Event B = Bob was late

We need to find P(A/B) = P(Bob drove to work, given Bob was late)

By Bayes' Theorem,

P(A/B) = P(B/A)P(A) / P(B)

P(A) = P(Bob drove to work) = 10% = 0.1

P(B/A) = P(Bob was late given he took the car) = 50% = 0.5

P(B) = P(Bob was late) = P(B/A)P(A) + P(B/A')P(A')

where A' = Bob not use the car = Bob took the bus or train

P(Bob took the train) = 90% = 0.9

P(Bob took the bus) = 0 (As he never takes the bus)

P(B/A')P(A') = P(bob was late given he took the train)P(bob took the train) +  P(bob was late given he took the bus)P(bob took the bus)

P(B/A')P(A') = 0.01(0.9) + 0.2(0)

P(B/A')P(A') = 0.01(0.9)

P(B) = 0.5(0.1) + 0.01(0.9)

Plugging in all the values --->

P(A/B) = P(B/A)P(A) / P(B)

P(A/B) = 0.5(0.1) / 0.5(0.1) + 0.01(0.9)

P(A/B) = 0.05 / 0.05 + 0.009

P(A/B) = 0.05 / 0.059

P(A/B) = 0.847

Ans : The coworkers probability that Bob drove to work that day, given that he was late = 0.847

Cheers!

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