Consider the Monty Hall problem, except that Monty enjoys opening door 2 more than he enjoys opening door 3, and if he has a choice between opening these two doors, he opens door 2 with probability p, where 1 2051 To recap: there are three doors, behind one of which there is a car (which you want), and behind the other two of which there are goats (which you don't want). Initially, all possibilities are equally likely for where the car is. You choose a door which for concreteness we assume is door 1. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the car, will always open a goat door and offer the option of switching, and as above assume that it Monty Hall has a choice between 1 opening door 2 and door 3, he chooses door 2 with probability p (with sp51), Hint: A prior knowledge which is not shown in the basic events is that you choose door 1 in the first step. I don't write it down explicitly to make the formula look simpler. Please keep it in mind that you choose door 1 in the first step. (b) (4 pts) Find the probability that the strategy of always switching succeeds, given that Monty opens door 2. Answer: 1/(1+p) 1) Define basic events: E2: Monty opens door 2 Ci: the car is behind door i. 2) Extract probability information. (P(CI), P(E2IC1). P(E21C2). P(E21C3)?) 3) Reasoning We choose door 1, Monty opens door 2, and then we switch to door 3. We win it and only if the car is behind door 3. Therefore, our wining possibility is P(C3|E2) 4) Compute P(C31E2). Use Bayes' rulo and LOTP (use C1, C2, and C3 as the partition of sample space) to solve it. (c) (3 pts) Find the probability that the strategy of always switching succeeds, given that Monty opens door 3.