Probability question involving a lift

[tantrik]

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 has stopped 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 though it 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 continuing to rise from first floor and start descending from the fourth floor? Let me know where the mistake is.

Thanks is advance.

 

[MarkFL]

I agree with your methods and results for parts a) and b). For part c), we could observe that for 4 stops, we need 2 of them to go up and 2 of them to go down in order to return to the point of origin. And then we need to look at the number of ways these 4 stops can be arranged. More simply, what we can use is binomial probability formula:

$$P(x)={n \choose x}p^x(1-p)^{n-x}$$

Here, $n$ is the number of trials, or stops in our case which is 4. We may take $x$ to either be the number of ups or the number of downs, and in our case these are both 2. If we take $x$ to be the number of ups, then $p$ is the probability of going up from a stop, which is $$\frac{3}{4}$$

Can you proceed?

 

[Opalg]

For part c), we could observe that for 4 stops, we need 2 of them to go up and 2 of them to go down in order to return to the point of origin. And then we need to look at the number of ways these 4 stops can be arranged. More simply, what we can use is binomial probability formula:

$$P(x)={n \choose x}p^x(1-p)^{n-x}$$

Here, $n$ is the number of trials, or stops in our case which is 4. We may take $x$ to either be the number of ups or the number of downs, and in our case these are both 2. If we take $x$ to be the number of ups, then $p$ is the probability of going up from a stop, which is $$\frac{3}{4}$$

Can you proceed?

Not sure I entirely agree with you there, Mark. Starting from floor 1, you can go down (D) or up (U). To finish where you started from, after four moves, you certainly need two Us and two Ds. But they cannot come in any order because there is only one floor below floor 1 – the office block apparently has no basement. So the sequence DDUU is not possible. (But any other sequence consisting of two Ds and two Us is allowed.)

A further complication is that whenever you reach the ground floor, the next move must be U, and in that case it occurs with probability $1$ rather than $3/ 4.$

I think that the safest way to get the answer is to list all the possible sequences of moves, namely UUDD, UDUD, UDDU, DUUD, DUDU. Then calculate the probability for each of them and add the results.

 

[MarkFL]

Yeah, I realized while I was out that I had goofed on this. I need to just stay away from probability problems. :)

 

[tantrik]

Not sure I entirely agree with you there, Mark. Starting from floor 1, you can go down (D) or up (U). To finish where you started from, after four moves, you certainly need two Us and two Ds. But they cannot come in any order because there is only one floor below floor 1 – the office block apparently has no basement. So the sequence DDUU is not possible. (But any other sequence consisting of two Ds and two Us is allowed.)

A further complication is that whenever you reach the ground floor, the next move must be U, and in that case it occurs with probability $1$ rather than $3/ 4.$

I think that the safest way to get the answer is to list all the possible sequences of moves, namely UUDD, UDUD, UDDU, DUUD, DUDU. Then calculate the probability for each of them and add the results.

Great many thanks for your in-depth explanation towards the question. I got the correct answer this time. Especially this info has been the most useful -

A further complication is that whenever you reach the ground floor, the next move must be U, and in that case it occurs with probability 1 rather than 3/4.

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