Number 2 please help quick .Asap

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

P(A and B) = P(A)*P(B given A)

P(A and B) = (3/4)*(3/4)

P(A and B) = 9/16

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

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

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

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