There are 100 people in line to board a plane with 100 seats. The first person has lost his boarding pass, so he takes a random seat.
Everyone that follows takes their assigned seat if it's available, but otherwise takes a random unoccupied seat. What is the probability the last passenger ends up in his/her assigned seat, as a decimal?
To calculate the probability that the last passenger ends up in his/her assigned seat, let's analyze the situation step by step.
1. First, we consider the special cases:
a. If the first person (passenger 1) happens to sit in his assigned seat (seat 1), then all the subsequent passengers will sit in their assigned seats. In this case, the last person will definitely end up in his/her assigned seat.
b. If passenger 1 sits in passenger 100's seat, then all subsequent passengers will take their seats, including passenger 100. Again, the last person will end up in his/her assigned seat.
Therefore, in these two cases, the last passenger will always end up in his/her assigned seat with a probability of 2/100 (since there are 2 possible outcomes).
2. In all other cases, passenger 1 will sit in a seat that doesn't belong to him or the last passenger. Let's consider the possible scenarios:
a. If passenger 1 sits in seat 2, then passenger 2 would have to sit in a random unoccupied seat to avoid taking passenger 1's seat. In this case, there are two possibilities:
- Passenger 2 sits in passenger 100's seat. In this scenario, all subsequent passengers would take their assigned seats, including passenger 100. The last person would end up in the correctly assigned seat.
- Passenger 2 sits in any other random unoccupied seat. In this case, all subsequent passengers would take their assigned seats until passenger 99, who would have two choices: their assigned seat or passenger 100's seat. If passenger 99 sits in their assigned seat, then all subsequent passengers would sit correctly and the last person would end up in the correctly assigned seat. If passenger 99 sits in passenger 100's seat, the chain reaction starts again, and we would end up in Scenario b.
b. If passenger 1 sits in seat N (where N ≠ 2, N ≠ 100), then passengers 2 through N-1 will take their assigned seats. Passenger N would have two options: their assigned seat or passenger 100's seat. If passenger N sits in their assigned seat, all subsequent passengers to passenger N - 1 will sit in their assigned seats, and the last person would end up in the correctly assigned seat. However, if passenger N sits in passenger 100's seat, we would revert to Scenario a.
Based on the above analysis, we can conclude that the only scenarios where the last passenger doesn't end up in his/her assigned seat are the two scenarios described in Step 1, and the Scenario b derived from Scenario a in Step 2.
Therefore, the probability that the last passenger ends up in his/her assigned seat is equal to 1 minus the sum of the probabilities of these three cases.
Probability = 1 - (2/100 + Probability of Scenario b)
To calculate the probability of Scenario b, we need to consider that passenger 1 has N-1 options for a seat (where N is the total number of seats), and passenger N has two options for a seat. All other passengers would sit in their assigned seats.
Probability of Scenario b = (N-1)/N * 1/(N-1) = 1/N
Substituting this into the equation:
Probability = 1 - (2/100 + 1/100)
Probability = 1 - 3/100
Probability = 97/100
Therefore, the probability that the last passenger ends up in his/her assigned seat is 97/100, or 0.97 as a decimal.