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alexandreC
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Joined: June 9th, 2004, 11:35 pm

Airplane Passengers

December 23rd, 2004, 4:53 pm

well, i dont think there is much left to say, Zerdna, invoquing a symmetry of the problem, presented a beautifull solution for half of the thread,Aaron in turn, rewrote P(n) and gave a very elegant interpretation for the other half!Aaron & Zerdna, thank you for the Christmas presents!Happy Christmas to all!!Alex
 
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jiantao
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Joined: November 23rd, 2004, 10:11 pm

Airplane Passengers

December 25th, 2004, 2:06 am

Suppose that there are total N passengers. The probability that the first passenger is not properly seated is: The probability that the second passenger is not properly seated is: The probability that the third passenger is not properly seated is: The probability that the K th passenger is not properly seated is: Therefore, the probability that the last passenger is properly seated is: 1/(N-N+2)=1/2There expected number of passengers who are NOT properly seated is: (N-1)/N+1/N+1/(N-1) +1/(N-2) +...+1/2
Last edited by jiantao on December 24th, 2004, 11:00 pm, edited 1 time in total.
 
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alexandreC
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Joined: June 9th, 2004, 11:35 pm

Airplane Passengers

December 25th, 2004, 11:47 am

(jiantao, that is all true, but Aaron has said that already...)
Last edited by alexandreC on December 24th, 2004, 11:00 pm, edited 1 time in total.
 
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citiboy
Posts: 14
Joined: January 12th, 2011, 5:28 pm

Re: Airplane Passengers

December 9th, 2020, 8:45 am

I believe the answer is 1/100

Reasoning: 

The first person will take a random seat which will not be the seat allocated to the last person with probability 99/100
The second person will take a random seat out of 99 remaining which will not be the seat allocated to the last person with probability 98/99

And so on, until the last but one person comes and there are two seats available. That person will not take the seat allocated to the last person with probability 1/2

Multiplying these probabilities we get 99/100 * 98/99 * 97/98 * ... * 2/3 * 1/2 = 1/100