Question

If you can explain your reasoning that will be appreciated.

You have been trapped in a building with 3 hallways, and each of them can lead to freedom.

Unfortunately for you, each of these 3 hallways has 2 doors. Each of the six doors could be

locked or unlocked independent of any other door. To be able to escape to

freedom through a particular hallway, you require that both the doors in that hallway are

unlocked.

The doors in hallway 1 is independently locked with probability 1/2

The doors in hallway 2 is independently locked with probability 1/3

The doors in hallway 3 is independently locked with probability 1/4

What is the probability that hallway 1 was open (in essence, both doors in hallway A were

unlocked) given that you managed to escape to freedom? Assume that if a hallway is closed you can go back and try another one.

Answer 1

P(managed to escape to freedom) = P(managed to escape to freedom | hallway 1 was open) * P(hallway 1 was open) + P(managed to escape to freedom | hallway 2 was open) * P(hallway 2 was open) + P(managed to escape to freedom | hallway 3 was open) * P(hallway 3 was open)

                                                       = (1 - 1/2)² * (1/3) + (1 - 1/3)² * (1/3) + (1 - 1/4)² * (1/3)

                                                       = 0.4190

P(hallway 1 was open | managed to escape to freedom) = P(managed to escape to freedom | hallway 1 was open) * P(hallway 1 was open) / P(managed to escape to freedom)

                                                       = (1 - 1/2)² * (1/3) / 0.4190

                                                       = 0.1989 (ans)