Thursday, March 15, 2012

I lost my Boarding pass

I used to hear people saying that the big challenge in problem solving lies in understanding the problem.The solution just follows that... This puzzle is a classical example.


On a sold out flight, 100 people line up to board the plane. The first passenger in the line has lost his boarding pass, but was allowed in, regardless. He takes a random seat. Each subsequent passenger takes his or her assigned seat if available, or a random unoccupied seat, otherwise. What is the probability that the last passenger to board the plane finds his seat unoccupied?


One can make this very complex by starting from Passenger 1 to 100 and working out all possibilities.But this question is as simple as below,

What is the probability for a passenger to get a specific seat among 100 seats? It is a simple probability question is not it?

answer is : 99!/100! = 1/100 = 1%

n(A) = other 99 seats can be occupied by other passengers = 99! ways.

n(s) =  number of ways 100 passengers can occupy 100 seats =100!

No comments: