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The last passenger is a good example of probability well understood: the trap is not in the count, but in choosing poorly the space of cases that really matters.
An airplane has \(n\) seats and \(n\) passengers, each with their assigned seat. Passenger 1 loses his card and sits randomly in any seat. Starting with passenger 2, each one acts like this:
Answer:
$$ \mathbb{P}(\t\t\text{último en su asiento})=\tfrac12 \quad (n\ge2). $$ **Explanation:** While the process is still open, only two entries matter: **1** and **n**. Every time a passenger finds their seat occupied and has to choose at random, three things can happen: 1. take seat 1, and from that moment the last one will end well; 2. takes the seat $n$, and the last one will no longer be able to sit in his seat; 3. take any other, and the problem is transferred to the owner of that seat, without changing its structure. So it all ends when one of those two critical seats first appears. Since both are in a symmetric situation, each one occurs with probability $1/2$. That is why the probability that the last one sits in his place is exactly $1/2$. $$
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