Question

Bob can decide to go to work either by car, bus, or commuter train. If he decides to go by car, there is a 50% chance he will be late due to traffic. 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, there is a 1% chance that he is late, but is more expensive than the bus. What is the probability that Bob drove to work today given he arrives late to work?

Answer 1

The probability is 0.7043

Explanation:

Let's call:

C: Bob goes to work by car

B: Bob goes to work by bus

T: Bob goes to work by Train

L: Bob arrives late to work

The probability that Bob drove to work today given that he arrives late to work is:

P(C/L) = P(C∩L)/P(L)

Where P(L) = P(C∩L) + P(B∩L) + P(T∩L)

So, the probability P(C∩L) that Bob goes to work by car and arrives late to work is calculate as:

P(C∩L)=(1/3)*(0.5) = (1/6) = 0.1667

Because 1/3 is the probability that Bob choose the car and 0.5 is the probability that he will be late at work given that he choose a car.

At the same way, the probability P(B∩L) that Bob goes to work by bus and arrives late to work and the probability P(T∩L) that Bob goes to work by Train and arrives late to work are calculate as:

P(B∩L)=(1/3)*(0.2) = (1/15) = 0.0667

P(T∩L)=(1/3)*(0.01) = (1/300) = 0.0033

Then P(L) is calculated as:

P(L) = 0.1667 + 0.0666 + 0.0033

P(L) = 0.2367

Finally, The probability P(C/L) that Bob drove to work today given that he arrives late to work is:

P(C/L) = 0.1667/0.2367 = 0.7043