Assume RACV (Royal Automobile Club Victoria) holds a rewarding carnival in South Bank, Melbourne. They have an activity where a new car is parked behind one of three doors (door A, B, C). Players are going to choose one door. Suppose you choose door A. Then, the host will open a door with no car behind. At this time, the host will ask you if you want to change your choice. Are you going to change your mind? To answer this question, you need to calculate the probability of winning the car when you stick to your decision and the probability of winning the car when you change your mind. P.S. The host won't tell you if there is no car behind your selected door.
Can anyone explain this question, I can't think of a way to solve this
I know that I am are supposed to calculate P(Winning|Decision Changed) and P(Winning|Decision Not changed).
I can say, P(Winning)=1/3 maybe then what about P(Decison Changed) and P(Decison Not Changed)
Changing your decision is not a random event, it is a strategy.
Rather you should consider the event of the prize being behind your first choice.
If the prize is there, then you will certainly lose if you change, while if the prize is not there it will certainly be behind the remaining door (thus you win if you change).