# Monty Hall Probability Question- with what i have done so far

#### verygreen

##### New Member
In a game show, there are 3 doors. Behind one of the doors is a
car, behind the other two are goats. A contestant chooses door number 1, but doesn’t open it right
away. The game show host, who knows behind which door the car is located, opens one of the
other two doors, revealing a goat.

Given that:
Fj := {the car is behind door j} (j = 1, 2, 3)
E := {the host opens door number 2}.

(a) Explain why P(F1) = P(F2) = P(F3) = 1/3.

(b) Explain why P(E|F1) = 1/2 and P(E|F2) = 0 and P(E|F3) = 1.

So far I have:

For A) I understand there is a 1/3 chance because there is 1 car, 3 doors.

For B)

For P(E|F1) = 1/2 - I think this is because the contestant has a 1/2 chance of deciding to switch with a 2/3 chance of winning and a 1/2 chance of staying with door 1 with a 1/3 chance of winning. Therefore, ( 1/2 x 2/3 ) + (1/2*1/3)=1/2.

Then for P(E|F2) = 0, there is a 0 chance because the host opened that door and there is a goat.

Then for P(E|F3) = 1 I am lost, I don't understand how to explain a probability of 1 (100%) when there is still a chance for the 1st door that was chosen.

Any help of direction would be greatly appreciated! Thanks

#### Mean Joe

##### TS Contributor
Then for P(E|F2) = 0, there is a 0 chance because the host opened that door and there is a goat.

Then for P(E|F3) = 1 I am lost, I don't understand how to explain a probability of 1
P(E|F2)=0 because the host will not open Door 2 if the car is behind Door 2.

P(E|F3)=1 because, recall that the contestant picked Door 1, AND the host will show either Door 2 or Door 3.