Probability question involving a lift

In summary, a lift in an office block has a 3/4 probability of continuing to rise after stopping on any floor (except the fourth floor) and a 1/4 probability of continuing to go down after stopping on any floor (except the ground floor). For a lift currently at the first floor and having just descended, the probability of its second stop being the third floor is 9/16 and the probability of its third stop being the fourth floor is 27/64. To return to the first floor as its fourth stop, there must be 2 stops going up and 2 stops going down, with a total of 4 stops. The safest way to get the answer is to list all possible sequences of moves and
  • #1
tantrik
13
0
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.
 
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  • #2
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:

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

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 \(\displaystyle \frac{3}{4}\)

Can you proceed?
 
  • #3
MarkFL said:
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:

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

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 \(\displaystyle \frac{3}{4}\)

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.
 
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  • #4
Yeah, I realized while I was out that I had goofed on this. I need to just stay away from probability problems. :)
 
  • #5
Opalg said:
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.
.
 

FAQ: Probability question involving a lift

What is the likelihood of getting stuck in a lift?

The probability of getting stuck in a lift depends on various factors such as the age and maintenance of the lift, the number of people inside, and the duration of the ride. Generally, the probability is very low, but it can increase in older or poorly maintained lifts.

What is the probability of the lift stopping at a specific floor?

The probability of the lift stopping at a specific floor depends on the number of floors in the building and the number of people inside the lift. If there are more people, the lift is more likely to stop at different floors to pick them up or drop them off, reducing the probability of it stopping at a specific floor.

What are the chances of a lift malfunctioning and causing an accident?

The probability of a lift malfunctioning and causing an accident is very low. Modern lifts are equipped with various safety features and undergo regular maintenance to ensure they are functioning properly. However, it is always important to follow safety guidelines and report any issues with the lift immediately.

How does the weight of the lift and its occupants affect its probability of functioning properly?

The weight of the lift and its occupants can affect its probability of functioning properly. Lifts have weight limits that are determined by the manufacturer, and exceeding this limit can increase the chances of a malfunction. It is important to follow weight restrictions and not overload the lift for safety reasons.

What is the probability of being in a lift with someone who is infected with a contagious disease?

The probability of being in a lift with someone who is infected with a contagious disease is difficult to determine as it depends on various factors such as the number of people using the lift, the prevalence of the disease in the area, and the precautions taken by individuals. It is always important to follow hygiene and safety guidelines to reduce the risk of spreading diseases in public spaces like lifts.

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