Problem:
You and 4 more friends bought tickets to watch a Rock n' Roll concert in another city. However, neither of you guys combined which seats were buying. That is, the tickets were bought randomly. You just noticed this fact while entering on the boarding queue.
Knowing that the plane model Boeing 737-700 has 102 seats at the economic class. And the seats are disposed in 2 lines, 3 per 3 seats. Having a corridor separating the lines.
I'm not quite sure what this means. Do you mean that the 102 seats have 102/2= 51 seats on each side of the isle, each row on one side of the aisle have 3 seats so there are 51/3= 17 rows (or 102/6= 17)?
which are the probabilities:
-to all of you are separated ( are in different rows)?
-all of you together ( same row)?
-the probability to you sit at least with one friend ( a friend in the same row)?
In order that more than 3 (so all 5 of you are together) you must be considering seats on both sides of the isle to be 'in the same row' so there are 17 rows. The first person is in
some row. That leaves 16 other rows so the probability the next person is in a different row is 16/17. There are then 15 rows left so the probability the next person is in yet another row is 15/17. Similarly we get 14/17, 13/17, and 12/17. The probability all 5 are in different rows is (16/17)(15/17)(14/17)(13/17)(12/17).
For the probability that all 5 are in the same row, the first person sits in
some row. There are 5 seats left in that row as opposed to 101 unoccupied seats on the plane. The probability the next person sits in one of the 5 seats left in that row is 5/101. There are then 4 seats left in that row out of 100 unoccupied seats. The probability the third person sits in that same row is 4/100. Similarly the probability the fourth person sits in that row is 3/99 and the probability the fifth person also sits in that same row is 2/98. The probability that all 5 sit in the same row is (5/101)(4/100)(3/99)(2/98).
I thought, for a moment, that the last would be just the opposite of the first, 1- (16/17)(15/17)(14/17)(13/17)(12/17). But that would be the probability that some two people sit in the same row. The probability that
you, personally share a row with at least one other person is different. But it is still simpler to find the probability that do NOT share a row with anyone and subtract from 1. You are in some row. the next person can sit in any of the other 6*16= 96 seats in the other 16 rows. The probability of that is 96/101. The third person can sit in in of the 95 seats remaining in the other 16 rows (this is different from the first question in that the third person
can sit in the same row as the second person). The probability of that is 95/100. Similarly the probabilities the fourth and fifth persons do not sit in the same row as you are 94/99 and 93/98. The probability that
you do
not share a row with a friend is (96/101)(95/100)(94/99)(93/98). The probability that you share a row with at least one friend is 1- (96/101)(95/100)(94/99)(93/98).
