Question

Now we modify it so that you are given the choice of four doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what is behind the doors, opens two other doors, say #3 and #4, each of which has a goat. Here we assume that the host cannot open the door to expose the car and when he can open two out of three doors, he chooses them at random. What is the probability of winning the car if you switch your choice and pick door #2?

(1) 1/4

(2) 1/3

(3) 1/2

(4) 3/4

Answer 1

P(opens 3 and 4)= P(car behind 1 and opens door 3 and 4)+P(car behind 2 and opens door 3 and 4)

=(1/4)*(1/C(3,2))+(1/4)*(1)

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

therefore P( winning the car if pick door #2)=(1/4)*(1)/(1/3) =3/4