In: Math
Here in gist , the sum is there are three doors one with a car and another two with goat. A guest choose a door. Then Host opens one of the other two door and bring out a goat. Then the guest get another chance to switch the door and the contestant switches. So we have to find the probability that the contestant wins a car.
Case 1. The host knows where the car and always opens a door showing a goat.
In that case if (x,y) be the outcome where x be the first choice and y be the switched choice then the sample space be,
{ (car, goat) , (goat , car) } since the host always opens a door with gaot knowingly hence after first choice there is always an alternative choice.
Hence the probability of getting a car after switching is 1/2.
Case 2. The host does not know where the car and opens a door showing a goat under random chance.
Then let us assume A to be the event that the guest initially chooses a goat, B to be the event that the host randomly chooses a goat door to open. Then there might be a chance that the host opens the door with car.
Then P(A) = 2/3 ( 2 out of 3 gates contains goat )
So P() = 1-2/3 = 1/3
P(B|A) = 1/2 ( after choosing a door with goat the host will choose a door with goat with probability 1)
and =1 ( after choosing a door with car the host will surely a door with goat)
So the probability that if the guest switches he will win a car
=P(A|B) ( since after randomly choosing a agate with goat by host the guest switches and win a car is equivalent to that the guest chooses a gate with goat at first)
(by Bayes theorem )
Hence the answer.............
Thank you.............