Lost boarding pass

Number the people 1, 2, 3, and so on, so that 1 boards first, 2 boards next, and 100 boards last.

First of all, observe that for $2 \le n \le 100$ , * *n 's seat is always taken after she finishes boarding.**. This is clear since if it isn't taken when she boards the plane, she will take it.

It follows that when 100 is getting on the plane, seats 2 through 99 are taken. Therefore, 100 will either get his own seat, or he will get 1's seat. That is, $P(\text{100 gets his seat}) + P(\text{100 gets 1's seat}) = 1 \quad \quad \tag{*}$

However, 1's seat and 100's seat are indistinguishable for the first 99 passengers . If any of them are to take either 1's seat or 100's seat, they pick randomly between the two. Therefore, the chance that one of the first 99 passengers takes 1's seat is equal to the chance that one of them takes 100's seat. But one of the first 99 passengers takes a given seat if and only if 100 does not get that seat. So we have $P(\text{100 doesn't get seat 1}) = P(\text{100 doesn't get seat 100})$ which is the same as $P(\text{100 gets his seat}) = P(\text{100 gets 1's seat})$ So from $(*)$ , $P(\text{100 gets his seat}) = \boxed{\frac{1}{2}}.$

Last person will either find the allotted seat for him or the seat for the first person vacant. Any other seat can't be vacant as if it was the case , it would have been already occupied by the real user.

There are two chances, geting assigned seat or not, Therefore p(geting assigned seat ) = 1/2 = .5

Asume that 1 is the value of a person get his right seat, otherwise, the value is 0 (ex. 00...00 is the case that nobody get his right seat. 111..111 is the case that everyone get his seat). There 100 people ==> 2^100 possibility. So the possibility for the final person get his right seat is: 2^99/2^100 = 1/2 ('cause the final person's value is 1, first 99 persons can be 1 or zero)

let us first calculate total cases such that last passenger's seat is occupied by someone else

if 1st passenger takes last passenger's seat ie 1 way if 1st passenger takes the seat of 99th passenger then 100's seat can be occupied by 99th passenger ie 1 way

if 1st passenger takes the seat of 98th passenger then 100's seat can be occupied in 2 ways(by 99th or by 99th)

if 1st passenger takes the seat of 97th passenger then 100's seat can be occupied in 4 ways(97th pass. occupies 98th place and then 100th place can be occupied in two ways as above,97th occupies 99th's place and 97 occupies 100's place)

similarly if 1st passenger takes 2nd seat then total no. of ways such that 100th passenger seat is taken by someone else is 2^97

total ways in which last passenger does not get his seat is

2^97 + 2^96 + ........ 2 + 1 +1 =

2^98

to calc. total possibilities

if 1st passenger takes 100th's seat then 1 way

if 1st passenger takes 99th's place then 2 ways(ie 99th will take 100th's seat or 1st seat ;others will take their own seats)

similarly if 1st passenger takes 2nd seat then total ways are 2^ 98

if 1st passenger takes his own seat ,then seats are occupied in 1 way. thus total ways are

2^98 + 2^97+........2+1+1

= 2^99

therefore req. probability is

1-(2^98/2^99) =

0.50

For $1 <= i <= 98$ , passenger $i$ has three options:

1. He/she chooses the first passenger's seat. Chance: $1/r$ , where $r=101-i$ equals the number of remaining seats. This option implies succes for the last passenger.

2. He/she chooses the last person's seat. Chance is also $1/r$ . Implies failure.

3. He/she chooses another seat, say $j$ 's seat. Then the procedure advances with passengers seating their designated seats (optionally none) until passenger $j$ 's turn. Chance is $\frac {1}{r-2} * P_j$ for every $j$ ; where $P_j$ equals the chance of succes if it is passenger nr $j$ 's turn and both the first passenger's seat and the last passenger's seat are still unseated.

This algorithm either ends preemptively with success or failure (options 1 and 2; chances of success and failure are equal), or ends with passenger $99$ who has two seats left with also 50% chance of success.

It's Just simple. Last person may get his own seat or any other seat with equal probability.So, ans is 0.5