On a sold out flight, 100 people line up to board the plane. The first passenger in the line has lost his boarding pass, but was allowed in, regardless. He takes a random seat. Each subsequent passenger takes his or her assigned seat if available, or a random unoccupied seat, otherwise. What is the probability that the last passenger to board the plane finds his seat unoccupied?

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Cheers,
:) James:)

When the last passenger boards the plane, there are just two possibilities: the one remaining seat may be his or that of the first passenger. Under the assumption that no preference has been exhibited by the boarding passengers towards either of the two seats, they both have the same probability to become the last unoccupied seat: 50%

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suresh

Hi,

I was also thinking like you. But it looks like the answer is wrong... [].

But now I am clear with the right approach and i know the answer [].

I would like to give one more chance to you all to find the right answer[]

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Cheers,
:) James:)

Well i guess its 99%.

for the second person who gets in has 1% possibility of his seat being occupied cz there is only one person who might sit there.

For the third person, there is 2% of possibility of his seat being occupied cz there are 2 people already there.

Likewise for the last person, it's 99% because there are 99 people and any of them can be occupied.

Is it correct???




_Niki_

nice q? yar .... can u tell me ans?

hi friend..

As per my view let the first passenger to wait until all the passengers with boarding pass are seated ...finally one seat will be empty and its easy to find out the seat of the passenger who lost his pass...

I have provided the answer because of the following line

"The first passenger in the line has lost his boarding pass, but was allowed in, regardless"

here it is stated that passenger is allowed in but no idea whether he was seated or not..


james if i am wrong then tell me where i am wrong

Thanks
Deepasree

Deepa,

When the last passenger boards the plane, there are just two possibilities:

1.The one remaining seat may be his or that of the first passenger.

2.Under the assumption that no preference has been exhibited by the boarding passengers towards either of the two seats, they both have the same probability to become the last unoccupied seat: 50%.

3. 98 * 1/2 * 1/100 + 1/100 = 1/2

Thanks

Gokulakrishnan K

hi friend..

The first person is the person who lost his boarding pass... based on that i can't get your point "When the last passenger boards the plane, there are just two possibilities:".. all the passengers are holding the boarding pass except the first

Thanks
Deepasree

Deepa,

99 people had boarding pass.

only one person is not having the pass.

what is the probable for only one seat.

i.e either the seat is his or the lost one.

So the P(event) = 1/2

Probabilty of any event is always 1. Say P(A)

Because it may happen or not.

P(A)+P(-A)=1

Thanks

G K

hi,

a very detailed explanation

Thanks
Sushma

hi
As its a sold out flight, with 100 seats or say 100 passes there is no probabibility that the last or first occupant finds it unoccupied ..

correct me if iam wrong....

thanks
Simrran