A lift starts with 5 passengers and stops at 8 floors. Find the probability that no two or more passengers leave at the same floor. Assume that all arrangements of discharging passengers have the same probability.

Answer:- The probability that no two or more passengers leave at the same floor is $\frac{6720}{32768}.$The various arrangements of discharging the passengers to any one floor may be denoted by symbols like (3,1,1) to be interpreted as the event that three passengers leave together at a certain floor, one passenger at another floor and finally the last one passenger at still another floor. So there are seven possible arrangements ranging from (5) to (1,1,1,1,1). Now calculate the total probability by adding probability for each arrangement. E.g. In how many ways all the five passengers gets off at any one floor. The answer is in $\binom{8}{1}$ ways. In this way,you can arrive at my final answer $\frac{6720}{32768}$

  • $\begingroup$ Please add [self-study] tag (check stats.stackexchange.com/tags/self-study/info) $\endgroup$ – Tim Jun 26 '16 at 14:30
  • $\begingroup$ Because the answer depends on how to tell when "arrangements for discharging passengers" are distinct, could you please explain what that means? Consider two passengers on a three-floor elevator. The "arrangements" might constitute all $3\times 3=9$ distinct choices the passengers could independently make, each with chance $1/9$. Or, they might constitute just the six distinct counts of discharged passengers at the floors: three ways for two to get off at once and three ways for two to get off on different floors, each with chance $1/6$. Obviously the probabilities differ in the two cases. $\endgroup$ – whuber Jun 26 '16 at 16:13
  • $\begingroup$ @whuber, This is a sample without replacement. In other words, once the person gets off the elevator, he is not put back in the sample. We need to find the probability corresponding to having 4 passengers on the board selecting a separate floor and the last person having the choice of selecting one out of remaining four floors. $\endgroup$ – Dhamnekar Winod Jun 26 '16 at 16:26
  • $\begingroup$ I don't see how that interpretation applies to your problem. Since the passengers are choosing the floors (rather than the floors choosing the passengers), the "sample" seems to come from the eight floors. If there is no possibility of replacement, then by assumption it's impossible to get two of the same floor (and the answer is 100%). If you instead view this process as somehow "sampling" the people, then you need to include some additional mechanism to keep track of which floors were involved. $\endgroup$ – whuber Jun 26 '16 at 16:30
  • $\begingroup$ @whuber, I have edited my posted question along with answer. $\endgroup$ – Dhamnekar Winod Jun 27 '16 at 15:22

I'm not able to comment and this is my first time answering a self-study so let me know if changes are needed.

My interpretation of this problem is "all arrangements of discharging passengers" refers to all possible outcomes i.e. Passengers A, B, C, D, E getting off on (5th floor, 5th floor, 2nd floor, 3rd floor, 7th floor) or (3rd, 4th, 5th, 6th, 7th) or (1st, 1st, 1st, 1st, 1st), etc.

The problem is looking for P(no more than 1 passenger per floor) = (outcomes that meet the criteria / all outcomes). The outcomes that meet the criteria is a without replacement combinatorics problem.

  1. You can not replace the passengers but you can replace the floors in the non-criteria meeting instances of more than one passenger exiting on a given floor e.g. (1st, 1st, 1st, 1st, 1st).

  2. My interpretation is everyone will get off the elevator. I think the inclusion of the no passenger gets off any floor is not in the spirit of the question.

  3. Outcomes that meet that criteria $=$ instances where all the floors selected will be different.

Another hint/correction: You can select the 5 destination floors from the 8 available floors more than ${8 \choose 5}$ ways.


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