A game show has three doors for a contestant to choose from. Behind one of the...

A game show has three doors for a contestant to choose from. Behind one of the doors is a brand new car, and behind the other two doors are a goat. The contestant randomly
picks a door (⅓ probability of each). The host opens one of the other two doors and reveals a goat. The host then gives the contestant the option of keeping their original
door, or switching to the other unopened door. If the contestant switches, what is the probability of winning the car? explain why, to win the car, switching gives you two times higher

probability than not switching?

****5 to 7 sentences please.Thanks

Solutions

Expert Solution

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.............

milcah answered 9 months ago

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