100 people seat at random on a plane with 100 seats while they also have a ticket with an assigned seat , what is the probability to seat on your assigned seat if you are the 34th to get on the plane?

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Braylon Perez · Accepted Answer

From a probabilistic point of view, the event that I, as the 34th passenger, sit in my correct seat is the same as the event that I sit in my correct seat, and nobody before me sits in my seat. Let Ai be the event that passenger i does not sit in my seat, and B be the event that I do sit in my seat. Then we want:  P  (      A    1    ∩      A    1    ∩  ⋯  ∩      A      33    ∩  B  )  .We can expand this as  P  (      A    1    )  P  (      A    2    ∣      A    1    )  ⋯  P  (      A      33    ∣      A    1    ∩  ⋯  ∩      A      32    )  P  (  B  ∣      A    1    ∩  ⋯  ∩      A      33    )  .Since there are 100 seats and 99 of them are not mine,  P  (      A    1    )  =  99  /  100Given       A    1   (i.e., passenger 1 did not sit in my seat), there are now 99 unoccupied seats and 98 unoccupied seats that are not mine, so   P  (      A    2    ∣      A    1    )  =  98  /  99.Similarly,   P  (      A    3    ∣      A    1    ∩      A    2    )  =  97  /  98, and so on. The 33rd person has 68 unoccupied seats to choose from, of which 67 are not mine, so   P  (      A      33    ∣      A    1    ∩  ⋯  ∩      A      32    )  =  67  /  68Finally, we need   P  (  B  ∣      A    1    ∩  ⋯  ∩      A      33   We are given that 33 people sat in seats that were not mine. So there are 67 unoccupied seats, one of which is mine, so this probability is 1/67. Putting it all together we have:       99    100    ⋅      98    99    ⋯      67    68    ⋅      1    67    =      1    100    .

100 people seat at random on a plane with 100

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