Problem 2, Part (a)
A = event elevator's 1st stop at the 2nd floor
P(A) = 3/4
Since the elevator needs to go up one floor
B = event elevator has its 2nd stop at the 3rd floor
P(B given A) = probability event B happens given event A happened
P(B given A) = 3/4
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P(A and B) = P(A)*P(B given A)
P(A and B) = (3/4)*(3/4)
P(A and B) = 9/16
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Answer: 9/16
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Problem 2, Part (b)
A = event elevator's 1st stop at the 2nd floor
P(A) = 3/4
B = event elevator has its 2nd stop at the 3rd floor
P(B given A) = 3/4
C = event elevator has its 3rd stop at 4th floor
P(C given (A and B)) = 3/4
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P(A & B & C) = P(A)*P(B given A)*P(C given (A and B))
P(A & B & C) = (3/4)*(3/4)*(3/4)
P(A & B & C) = 27/64
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Answer: 27/64
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Problem 2, Part (c)
A = event elevator's 1st stop at the 2nd floor
x = P(A) = 3/4
B = event elevator has its 2nd stop at the 3rd floor
y = P(B given A) = 3/4
C = event elevator has its 3rd stop at 2nd floor
z = P(C given (A and B)) = 1/4
D = event elevator has its 4th stop at 1st floor
w = P(D given (A & B & C)) = 1/4
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P(A & B & C & D) = x*y*z*w
P(A & B & C & D) = (3/4)*(3/4)*(1/4)*(1/4)
P(A & B & C & D) = 9/256
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Answer: 9/256