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Most likely outside the scope of anything you will see on

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Most likely outside the scope of anything you will see on [#permalink] New post 22 Jun 2007, 07:57
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Most likely outside the scope of anything you will see on the GMAT, but it is interesting and a good mental exercise nontheless...


One hundred people are in line to board Airforce One. There are exactly 100 seats on the plane. Each passanger has a ticket. Each ticket assigns the passenger to a specific seat. The passengers board the aircaft one at a time. GW is the fist to board the plane. He cannot read, and does not know which seat is his, so he picks a seat at random and pretends that it is his proper seat.

The remaining passengers board the plane one at a time. If one of them finds the seat is already taken, they will pick a seat at random. This continues until everyone has boarded the plane and taken a seat.

What is the probability that the last person to board the plane sits in their proper seat?
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 [#permalink] New post 22 Jun 2007, 08:47
Semi-educated guess here, let me know if I'm on the right track...

GW: 1/100
2nd person: 1/99
3rd person: 1/98
4th person: ...

1 - (1/100)(1/99)(1/98)... = ...?
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 [#permalink] New post 22 Jun 2007, 08:59
I did this problem in two ways: the simple way and the complicated way. Each yielded a different solution !

The simple way
-------------------
The first problematic person will create a wave of mess and randomness, in which case the last person will definately not sit in his own seat as the incorrect seating point will touch someone eventually. So the probability that the last person will sit in his correct seat, as it shows in his ticket, is merely the probability that the first person will randomly choose his own seat = 1/100 = 1 %

So the answer is 1 % <--- more likely to be correct for me


The complicated way
--------------------------
Assuming we're talking about 2 people only, then there is 50 % the first person, the one who doesn't read, will correctly select his seat, although at random. The second person will not effect any probabiliy number because he will select whatever seat that's left. This says that the last event does not count.

let's examine the case of 4 pessengers. The first person, who doesn't read, has a probability of selecting his correct seat of 0.25 . Now, the second person has a probability of 1/3 x 3/4 that he'll find his seat taken and thus will select another seat at random.

The 3rd person will have a probability of 1/2 x 1/3 x 3/4 of finding his seat being taken. the 4th and last pessenger will have a probability of 1/16 that will have his seat taken --> 15/16 probability that he'll sit in his own seat, as it shows in his ticket.

It looks like the probability that the last person will end up sitting in his seat is 1- (1/2^n) where n is the total number of pessengers/seats.

so for 100 pessengers, it is 1 - (1/2^100) ~ 100% !!!
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 [#permalink] New post 22 Jun 2007, 09:51
Both wrong... but Mishari was on the right track in part of his answer.
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 [#permalink] New post 22 Jun 2007, 09:52
which part: the simple or the complicated way ?
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 [#permalink] New post 22 Jun 2007, 10:09
I got I got it ..

it is 99/100 = 99%


I used a tree diagram for 3 people, then 4 people, then 5 people and realized the logic behind the pattern.

it is 1 - 1/n where n is the number of pessengers/seats.

Because 1/n the probability that the last person will sit in the wrong seat.
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 [#permalink] New post 22 Jun 2007, 10:24
Mishari wrote:
which part: the simple or the complicated way ?


Complicated, first paragraph
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 [#permalink] New post 22 Jun 2007, 12:04
Is it 1-1/10000 = 9999/10000=99.99%
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 [#permalink] New post 24 Jun 2007, 10:01
OA is 1/2.

Mishari, you were correct in the case for 2 people. By induction, you can extrapolate that out for any n number of people.
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 [#permalink] New post 25 Jun 2007, 21:44
Well .. here is the logic I used,

we just need the probability that the last guy sits in the proper seat - which means the 99 guys before him DO NOT sit in the last guy's allocated seat

now,

1st guy entering - probability that he will not sit in the last guys seat - 99/100

i.e he can sit nywhere in the 99 seats but for the last guys seat.

similarly 2nd - 98/99 ; 3rd - 97/98 and so on

Finally it will be 99!/100! = 1/100

how do we get 1/2 - what is it I am missing here ?
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 [#permalink] New post 25 Jun 2007, 22:31
grad_mba wrote:
Well .. here is the logic I used,

we just need the probability that the last guy sits in the proper seat - which means the 99 guys before him DO NOT sit in the last guy's allocated seat

now,

1st guy entering - probability that he will not sit in the last guys seat - 99/100

i.e he can sit nywhere in the 99 seats but for the last guys seat.

similarly 2nd - 98/99 ; 3rd - 97/98 and so on

Finally it will be 99!/100! = 1/100

how do we get 1/2 - what is it I am missing here ?


The problem starts with the first guy. According to the problem everyone else will sit in his seat if it is available.

So the probabilty of the first guy sitting on his on seat = Probability of the last guy getting his allocated seat = 1/2.

probabilty of the first guy sitting on his on seat = 1/2 (he either sits or does not sit. It is not necessary to consider where he sits if he picks the wrong seat.)

Lets not consider 100 seats for the time being. Let there be just 2 seats. Then, the probability of the last guy (2nd here) getting the seat is 1/2. Agree?

Now, consider 3 seats. If the first guy sits in the second guy's seat then only there will be a need to calculate the probabilty (other cases would represent absolute events - either the last guy gets his seat or he doesn't). And this probabilty will be 1/2.

By induction we can extend this for any number of seats and the probabilty will remain 1/2. :-D
  [#permalink] 25 Jun 2007, 22:31
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