My approach to Monty Hall problem using Baye's Theorem (CONDITIONAL PROBABILITY):

P(A|B) = P(B|A)P(A)/(P(B|A)P(A)+P(B|A')P(A'))

LET A=P(CAR), B=SWITCH (AFTER 1ST GOAT IS SHOWN) GIVEN THAT IT IS CAR/GOAT

FROM EACH OF THE DOORS, SAY THE DOOR WITH A GOAT BEHIND IT (G1):

P(CAR|SWITCH) =
P(SWITCH|CAR)P(CAR)/(P(SWITCH|CAR)P(CAR)+P(SWITCH|GOAT)P(GOAT)

P(SWITCH|GOAT) = 1/2 (50% CHANCE THAT ONE OF THE DOOR PICKED HAS A GOAT BEHINDAFTER HOST SHOWS THE 1ST GOAT) P(SWITCH|CAR) = 1/2 (OR THERE'S 50% CHANCE THAT ONE OF THE DOOR PICKED HAS A CAR BEHIND IT AFTER THE 1ST GOAT IS SHOWN) P(CAR) = 1/3 (PROB OF GETTING A DOOR WITH A CAR FROM THE BEGINNING)

AGAIN, WANT P(CAR|SWITCH) = (1/2 *1/3)/((1/2*1/3)+(1/2*2/3) = 1/3 (THE PROB FROM DOOR WITH G1 OF GETTING A CAR GIVEN THAT SWITCHING IS DONE)

FROM DOOR WITH G2, SIMILARLY, P(CAR|SWITCH) = 1/3

FROM DOOR WITH C, P(CAR|SWITCH) = 0. PROB OF GETTING A CAR GIVEN THAT SWITCHING IS DONE IS NEVER, SINCE SWITCHING FROM DOOR WITH CAR ALWAYS YIELDS A GOAT. P(GOAT) = 2/3 (COMPLEMENT OF P(CAR))

P(CAR|SWITCH) FROM DOOR 1 U P(CAR|SWITCH) FROM DOOR 2 U P(CAR|SWITCH) FROM DOOR 3 = 1/3 + 1/3 + 0 = 2/3

THANKS.

HELEN TSAI