One hundred people line up to board an airplane. Each has a boarding

Question

One hundred people line up to board an airplane. Each has a boarding pass with assigned seat. However, the first person to board has lost his boarding pass and takes a random seat. After that, each person takes the assigned seat if it is unoccupied, and one of unoccupied seats at random otherwise. What is the probability that the last person to board gets to sit in his assigned seat?

(A) 0.01
(B) 0.1
(C) 0.25
(D) 0.5
(E) 0.75

(A) 0.01
(B) 0.1
(C) 0.25
(D) 0.5
(E) 0.75

MathRevolution · Answer

The question is wordy and the number '100' given is just to make calculations complex.

The last person has only two options: Either he gets the right seat or he gets the wrong seat.

Hence 50% chances.

Answer D

IanStewart · Answer

That there are only two outcomes does not make the probability of each outcome 1/2. The probability you win the lottery is not 1/2 because you either win or lose. The probability is only 1/2 if you can establish that the two possible outcomes are equally likely. That turns out to be true here, but it would be trivially easy to chance the seating assignment rule in this question to make the answer different from 1/2.

In this question, if we call the first few people A, B, C, D, etc, then when B gets on the plane, then 1/100 of the time, A is in B's seat, and the rest of the time, B's seat is available. So there is a 99/100 chance B gets her assigned seat. Regardless, after the first two people board, B's assigned seat is occupied by someone, and one random seat, of the other 99, is occupied. So when C boards, there will be a 98/99 chance C can sit in his assigned seat. But once that happens, then the seats that were assigned to B and C are both occupied, and one random seat is occupied. So this continues -- D has a 97/98 chance to sit in her assigned seat, E has a 96/97 chance, and so on, until we reach the last person, who has a 1/2 chance.

I think one useful strategy to try for this question, if one gets stuck, is solving the problem with 3 passengers instead of with 100. Then you can itemize all the cases, and more easily see what's going on (and if you get the right answer doing that, you can extrapolate from there).

ScottTargetTestPrep · Answer

Solution:

To solve this problem, we can use a much smaller number for the number of people who are boarding the plane. So let’s say there are only two people boarding the plane with the 1st passenger’s assigned seat is A and the 2nd passenger’s assigned seat is B. If the 1st passenger, who lost his ticket, takes A as his seat, then the 2nd passenger will take B as his seat. In this case, the 2nd passenger sits in the correct seat. However, if the 1st passenger, who lost his ticket, takes B as his seat, then the 2nd passenger will have to take A as his seat. In this case, the 2nd passenger sits in the wrong seat. We see that if there are 2 passengers, the probability that the last passenger (i.e., the 2nd passenger) sits in the right seat is 1/2.

Now let’s say there are only three people boarding the plane, and the 1st passenger’s assigned seat is A, the 2nd passenger’s assigned seat is B, and the 3rd passenger’s assigned seat is C. There are 3 scenarios:

Scenario 1:  
If the 1st passenger, who lost his ticket, takes A as his seat, then the 2nd passenger and 3rd passenger will take B and C as their seats, respectively.

Scenario 2: 
If the 1st passenger takes B as his seat, then the 2nd passenger will take either A or C as his seat. If he takes A as his seat, then the 3rd passenger will take C, his assigned seat. However, if the second passenger takes C as his seat, then the 3rd passenger will take A as his seat, which is the wrong seat.

Scenario 3: 
If the 1st passenger takes C as his seat, then regardless of which seat the 2nd passenger picks, the 3rd passenger will be sitting in the wrong seat.

The probability that the last person (i.e., the 3rd person) sits in the correct seat is 1/3 from Scenario 1 and 1/3 x 1/2 = 1/6 from Scenario 2 and 0 from Scenario 3. Thus, the overall probability is 1/3 + 1/6 + 0 = 2/6 + 1/6 = 3/6 = 1/2.

As we can see, regardless of the number of people boarding the plane, the probability that the last person will sit in the correct seat is 1/2.

Answer: D

qadro · Answer

Tricky problem. i doubt this will be on GMAT ever.

Regardless,

1. If the first person sits on his own seat, then the last person sits on his seat for sure. 
2. If the first person doesn’t sit on his seat, then the last person either will seat on his own seat or will sit on the first persons seat. This may be unintuitive.

if the first person sits on #6, then 2-5 seat correctly. Then 6 either sits on #1’s original seat or sits say on 15. In the latter case 7-14 sit correctly and the issue gets delayed recursively. This happens until the last incorrectly seated passenger sits on the last seat then the only vacant seat is 1’s. The only other way is that an incorrectly seated passenger sits on 1’s seat, in that case all the following passengers sit correctly.

With this hope it is evident that for the last person, 1’s original seat and seat #100 are the only 2 options.

With all this symmetry must be brought into play. For every passenger after #1, they either sit on 1s original seat or another incorrect seat. Every following passenger has this 50:50 choice. So i think a majority of these guys can be removed from the calculation to reduce the 100 passengers to 3-4 only to make things simple. So it is 1, 2(rep for 2-99) and 100. Out of these 3 it is clear that there are 4 cases that pan out. In 2 100 sits on the right place and 2 where 100 sits on 1s original place. So the chances here seem 50:50.

devashish2407 · Answer

IanStewart  Thank you so much for the detailed explanation.

Just have 1 question, when C boards why are we not looking at the total number of available seats instead on 99.

There were 2 people who boarded the flight, thus C should only have 98 options to choose from.

This might be the very basic question, but really want to understand what ois going on here.

redranger6963 · Answer

This problem can be solved by symmetry. The key insight is that the last passenger will end up in either their assigned seat or the seat of the first passenger. If at any point a randomly selecting passenger chooses the first passenger's seat, the last passenger will definitely get their assigned seat. Conversely, if a randomly selecting passenger chooses the last passenger's seat, the last passenger will not get their assigned seat. Since each randomly selecting passenger is equally likely to choose any seat, the probability of choosing the first passenger's seat is equal to the probability of choosing the last passenger's seat. Therefore, by symmetry, the probability that the last passenger will get their assigned seat is 1/2.
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One hundred people line up to board an airplane. Each has a boarding

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