Q:

A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days’ travel. The second leads to a tunnel that returns him to his cell after 4 days’ travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .5, .3, and .2, what is the expected number of days until the prisoner reaches freedom? 1. Repeat problem, assuming the prisoner remembers previously chosen doors, and does not re-choose them. Assume the probabilities for the other doors are proportionally larger. 2. Repeat problem but now suppose there is another cell, and that door 1 takes him to the other cell after 2 days of travel. For the other cell, there are two doors, one of which leads to freedom after 3 days of travel, and the other leads back to the prisoner’s original cell after 3 days of travel; each door is equally likely.

Accepted Solution

A:
Answer:dont escape lolStep-by-step explanation:the answer is 2