Suppose an airline accepted 12 reservations for a commuter plane with 10 seats. They know that 7 reservations went to regular commuters who will show up for sure (p=1). The other 5 passengers will show up with a 50% chance, independently of each other.


a) The probability that more passengers who will show up for sure will show up than seats are available

b) find the probability that there will be empty seats

c) Let X be the number of passengers turned away. Find E(X)

any hints on how to start?


I threw together a truth table that I think answers your questions:

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  • I show all possibilities for the 5 undecided passengers. In the "over?" column I show whether or not that situation leads to turned-away passengers. There are six out of 32 for an 18.75% chance at least one person will be turned away.

  • In the "empty seats?" column I show true if the number of undecided passengers who show up is less than 3. That happens 16 times out of 32 for a 50% chance.

  • In the last column I show the expected number of passengers turned away as 0.21875. I'm not sure I calculated this right however, but this will get you most of the way there.

  • $\begingroup$ thanks, Is there an analytical way?. you can send the file excel ? $\endgroup$ – albert Oct 12 '15 at 20:41
  • $\begingroup$ sorry, why chances = 0.03125? $\endgroup$ – albert Oct 12 '15 at 20:49
  • $\begingroup$ @albert, Probably, no, and it's 1/32. $\endgroup$ – user2023861 Oct 12 '15 at 21:09
  • $\begingroup$ @albert, I should explain further. I chose a truth table to make my calculations because the number of possible outcomes is so small. If there were 500 passengers instead of 12, I would use a different method. I think this method makes the logic more clear. With 5 independent, binary variables, you have only 2^5 possible outcomes (32). (Also, I already deleted my spreadsheet, but it's so simple that you can easily recreate it) $\endgroup$ – user2023861 Oct 12 '15 at 21:25
  • $\begingroup$ Thanks, it was useful to me. is correct, I confirmed analytically $\endgroup$ – albert Oct 12 '15 at 21:32

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