Probability question involving a lift

tantrik said:

Dear friends,

I am stuck at the following probability problem. Will appreciate your help.

An office block has five floors (ground, 1, 2, 3 and 4), all connected by a lift. When it goes up to any floor (except 4), the probability that after it hasstopped it will continue to rise is 3/4. When it goes down to any floor (except the ground floor), the probability that after it has stopped it will continue to go down is 1/4. The lift stops at any floor it passes.

The lift is currently at the first floor having just descended. Calculate the probability of the following events:

a) its second stop is the third floor
b) its third stop is the fourth floor
c) its fourth stop is the first floor.

For part (a), I did this: P(first stop is the second floor)*P(second stop is the third floor) = (3/4)(3/4) = 9/16

For part (b), I did this:

P(first stop is the second floor)*P(second stop is the third floor)*P(third stop is the fourth floor)

= (3/4)(3/4)(3/4) = 27/64

Since the lift is going up and continuing to rise after it stopped at each floor from first floor, finding answers to part (a) and (b) was easy, even thoughit is unsure whether my logic is sensible or not. Somehow the answers for part (a) and (b) are correct according to the book. If my methods and answers are incorrect, let me know.

Now, I am really stuck at part (c). Not understanding how to solve the problem. How will the lift’s fourth stop be the first floor if the lift is continuingto rise from first floor and start descending from the fourth floor? Let me know where the mistake is.

Thanks is advance.

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Ishuda said:

Parts (a) and (b) are correct. Part (c) gets tricky since we can have several ways to get to the first floor at the fourth stop. First list the number of ways and compute the probability, i.e. stop at floors 2, 1, 2, 1 is (1)*(1/4)*(1)*(1/4)=(1/16), then add them up.

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