Consider the following variation of the Monty Hall problem, where in some situations Monty may not open a door and give the contestant the choice of whether to switch doors. Specifically, there are 3 doors, with 2 containing goats and 1 containing a car. The car is equally likely to be anywhere, and Monty knows where the car is. Let 0 p 1. The contestant chooses a door. If this initial choice has the car, Monty will open another door, revealing a goat (choosing with equal probabilities among his two choices of door), and then offer the contestant the choice of whether to switch to the other unopened door. If the contestant's initial choice has a goat, then with probability p Monty will open another door, revealing a goat, and then offer the contestant the choice of whether to switch to the other unopened door; but with probability 1 p, Monty will not open a door, and the contestant must stick with their initial choice.

The contestant decides in advance to use the following strategy: initially choose door 1. Then, if Monty opens a door and offers the choice of whether to switch, do switch.

(a) Find the unconditional probability that the contestant will get the car. Also, check what your answer reduces to in the extreme cases p = 0 and p = 1, and briefly explain why your answer makes sense in these two cases.

(b) Monty now opens door 2, revealing a goat. So the contestant switches to door 3. Given this information, find the conditional probability that the contestant will get the car.