Event:
S=success for last passenger to get his assigned seat.
1=case 1
2=case 2
Note that cases 1 and 2 are complementary.
Case 1: P(1)=1/100
If the first person chose (randomly) his assigned seat
Then all the rest of the passengers will also get their assigned seats.
=> P(S|1)=1
Case 2: P(2)=99/100
If the first passenger did NOT choose the right seat.
Then
case 2a: the seat occupied by first passenger is never taken by the next 98 passengers, so the last passenger will be left with the first passenger's assigned seat => P(S)=0
case 2b: if the assigned owner of the seat taken by the passenger boards the plane, then he will have to occupy another seat. His own assigned seat will always be occupied by another passenger, including the last one, so again P(S)=0.
Conclusion: P(S)=0 for either case 2a or case 2b, -> P(S|2)=0 for case 2.
Therefore, by the law of total probability,
P(S)=P(S|1)*P(1)+P(S|2)*P(2)
=1/100*1 + 99/100*0
=1/100
(i.e. if the first passenger happened to have chosen his own seat)
There are 100 people in line to board a plane with 100 seats. The first person has lost their boarding pass, so they take a random seat. Everyone that follows takes their assigned seat if it's available, but otherwise takes a random unoccupied seat. What is the probability the last passenger ends up in their assigned seat?
2 answers
Fancy answer. Too bad its wrong. The correct answer turns out to be 1/2.
0 answers