The answer is 1/2 or a 50% chance. Here's how:
Let p(x) denote the probability that the xth passenger finds his/her
seat occupied when coming on board the plane.
When the first person gets on board, his seat is definitely NOT
occupied. so...
p(1) = 0
The second person will find his seat occupied if the first guy took
it, and the probability of that is 1/100. so...
p(2) = 1/100
The third person will find his seat occupied if EITHER the first
guy occupied it OR the second guy occupied it. so...
p(3) = 1/100 + p(2).1/99
Arguing similarly, let's p(4) and p(5):
p(4) = 1/100 + p(2).1/99 + p(3).1/98
p(5) = 1/100 + p(2).1/99 + p(3).1/98 + p(4).1/97
You should now see a pattern emerging:
p(2) = 1/100
p(3) = p(2) + p(2).1/99
p(4) = p(3) + p(3).1/98
p(5) = p(4) + p(4).1/97
....
....
p(x) = p(x-1) + p(x-1).1/(100-x+2)
....
....
p(100) = p(99) + p(99).1/2
Infact, it's easy to see why we get this pattern. Think about what
happens when person 'x' boards the plane. That depends on what happened
when the (x-1)th person boarded the plane. EITHER the xth seat was
already occupied OR the (x-1)th person occupied the xth seat because
the (x-1)th seat was taken.
We know that the probability that the (x-1)th seat was occupied
when person 'x-1' got on board is p(x-1). THE KEY IS to realize
that the probability that the xth seat was occupied at that time
is also, infact, p(x-1). Why? Because every occupied seat had the
same likelihood of getting picked (because people chose their seats
at random if their seats were taken). If we add the two probability
in the EITHER/OR case in the preceding paragraph, we get the formula:
p(x) = p(x-1) + p(x-1).1/(100-x+2)
Anyway, the rest is easy.
Simplifying our final pattern, we get the following:
p(2) = (1/100)
p(3) = (100/99).p(2)
p(4) = (99/98).p(3)
p(5) = (98/97).p(4)
...
p(100) = (3/2).p(99)
Since each probability is > 0, we can safely multiply all the
terms on either side and we get:
p(2).p(3)...p(100) = (1/100)(100/99).(99/98).(98/97)...(3/2).p(2).p(3)...p(99)
Canceling numerators & denominators on the right side and p(2).p(3)...p(99)
on either side, we get:
p(100) = 1/2.
So, the probability that the 100th guy sits in his/her own seat
= 1 - p(100) = 1/2.
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