How to solve monty hall with 5 doors using bayes theorem? With 5 doors we get the following events. A: Probability of choosing a door with a car behind it: 1/5
B|A: Likelihood of monty opened door B if door A contains a car: 1/4 ? Im not sure about this one
How can I proceed now?
With 5 doors he will open 3 doors. Door A is the door I choose, door B is the one monty left closed.
How can I obtain the probability that the car is actually behind door B? That is if i switched from door A from door B.
The result supposedly is 4/15 but i cannot achieve using bayes theorem.
Can someone guide me in the right direction? Thanks
A: You are not at all interested in the probability that Monty chooses a particular door.   You are entirely concerned about the probability that another door hides the prise when given that Monty always reveals one which does not.
So the events of interest are the car is behind the first chosen door (A) and the car is behind the door you would switch to (C).
Well, if the prise is behind the first door, switching surely loses, else if it is not behind the A-door, then after Monty reveals another door where it is not, the prize is equally likely to be behind each of the three remaining doors.
Thus we evaluate the marginal probability that switching wins as:
$$\begin{align}\mathsf P(C)&=\mathsf P(A)\,\mathsf P(C\mid A)+\mathsf P(A^\complement)\,\mathsf P(C\mid A^\complement)\\&=\tfrac 15\tfrac 03+\tfrac 45\tfrac 13\\&=\tfrac 4{15}\end{align}$$

Note that the probability that staying wins is $3/15$, so this game still makes switching the better option, though only slightly.
