Passengers having seat numbers from 1 to 100 are to be seated in a flight. There are 100 seats only in the plane and the passengers enter the plane according to their seat numbers (passenger having seat number 1 enters 1st and so on). Initially the plane is fully vacant but the passenger 1 instead of taking his seat occupies some random seat. Rest of the passengers sit according to their seat numbers if it is vacant and occupy a random seat otherwise.
What is the probability that the 100th person sits at the seat alloted to him?
62But these are not constant.. that is what you have to prove!!
I dont see how that is so obvious! May be that i am missing some much simpler trick that is obviousl to you not me!!
but i dont think that is very likely!
62Think over it ramkumar!
Your probability of anyone occupying that seat is misplaced... Read my solution above.. it is on similar lines!
1is it not??? except d 100th guy someone has to take it kno....
62p(anyone takin seat 1)=p(1st person taking seat 1)*p(2nd person takin seat1)*....p(99th person taking seat 1)?????????
is this true???
Thnk over it!
1i ll explain u how i guessed it...
p(100th person taking 100th seat)=p(anyone occupying seat no 1)
because only if seat 1 is occupied, 100th person wil get 100th seat..
p(anyone takin seat 1)=p(1st person taking seat 1)*p(2nd person takin seat1)*....p(99th person taking seat 1)
=(1/100)*(1/99)*(1/98)*.......(1/2)
=1/100!
1i ll explain u how i guessed it...
p(100th person taking 100th seat)=p(anyone occupying seat no 1)
because only if seat 1 is occupied, 100th person wil get 100th seat..
p(anyone takin seat 1)=p(1st person taking seat 1)*p(2nd person takin seat1)*....p(99th person taking seat 1)
=(1/100)*(1/99)*(1/98)*.......(1/2)
=1/100!
1so.......if d first person takes 5th seat.........then till 4 everything is ok...but d 5 th guy can take any seat after tht ......if he takes d first seat then ther is no prob....the ques is finishd..........but if he takes sum other seat......then tht guy also will shift again......this process may continue untill sum1 sits in d 1st seat..........
1w8.......... i hav a confusion now........u may be right.........d xth person sits sumwher else.........he can sit anywhere........ and then d problem starts again...........ya its a tough and long problem......
1hey ramku hw is tht possible dude...........d first guy only sits randomly.........all others sit according to their numbers except tht person whose seat is occupied by d first person....he sits sumwher else...
1byah. if person 1 sits in seat no 100 then thres no possibility fr person no 100 to get hs seat...if person 1 sits somewhere else at x.. then every one upto x-1 ll get their alloted seats..... then d actual xth person ll take some other seat randomly.. prob of xth person choosin seat no 1 is 1/100-x ... after dat illl telll u in a while... plz try to bring as equations.....
1ya ive realised now........it isnt tht obvious after all......... for example........ wat if d third guy enterin d plane sits in d 4th seat???? i havent taken tht into consideration.....ive just assumed tht every passenger is a good boy/girl and theyll sit in their places and if sum1 occupies his place hell sit in d next one........hehehe
1wel wat i said was........ d person cannot sit in seat 1 nor in 100.......so there are 98 possibilities left(favourable outcomes)........its already given tht he doesnt sit in d 1 st seat..............so 99 overall outcomes...
so prob=98/99........................MIGHT BE WRONG..........
1So nishant, we basically excluding the first seat(seat no.1)..... and d first person shudnt sit in d 100th place....... so ther r 99 seats(excludin 1st one) .... so prob is 98/99....
62Swetha, i have given a very detailed solution.. pls read both my posts.. the solution i have given is in 2 parts
62suppose
P(k) is the probability that passenger no k sits on seat no 1
so, P(1)+P(2)+....P(100) =1 ..... (i)
again
P(r)={1-P(1)-P(2).... - P(r-1)} × 1/(100 - r + 1)
Also, P(1)=0 {it is already known that person 1 doesnt sit on seat 1}
P(2) = {1 - 0} × 1/99 = 1/99
P(3) = {1 - 0 - 1/99} × 1/98 = 98/99 × 1/98 = 1/99
P(4) = {1 - 0 - 1/99 - 1/99 } × 1/97 = 97/99 × 1/97 = 1/99
...
thus, by continuing, each P(i) = 1/99 ( 2<=i<=100 )
Thus P(100) = 1/99
Thus, probability that person 100 sits on seat 100 is 1-1/99 =98/99
Hence 98/99
I hope this solution is clear.
62The question seems very tough. But i think akhilesh's answer is correct.
He has not given any explanation. I will try to do that.
We should note that for the 100th passenger to sit at seat 100, it should mean that no passenger should sit on seat 1 in between! if someone sits of seat 1, everyone after that will sit on his seat!
This is the critical step to understand!
How this happens. Note: at any time, suppose, r passengers are in the aircraft, then only one seat less than equal to seat no r is unoccupied! This is a bit tricky to understand.
Suppose that passenger 1 sat on seat "k". So passengers 2 to "k-1" sat on their respective seats!.
So if no of passengers is less than equal to k-1, only one seat less than seat no k is unfilled (namely seat 1)
So when passenger no k comes, he will sit on some seat other than "k". If it is seat 1, then no seat less than equal to k will be unfilled!
If he sits on another seat, say seat L, even then only one seat less than equal to k is unfilled (namely seat 1)
now you can keep going this way forever. You will realise that if n passengers have entered the plane, at maximum, one seat less than equal to seat no n is unfilled. That too seat no 1!
So if someone sits on seat 1, we are done. So we have to find the probability of someone sitting on seat 1 before 100th no passenger comes!
