Thread: Statistical Analysis

Hi, This problem has been taken from the ' statistical analysis ' question paper for master of commerce examination of well-known university in India. "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." Ans: I don't know the correct answer. I expect some hint from the members of this mathhelpforum to solve this problem.I haven't used latex here

Hey Vinod.

For this problem the first you have to get the assumption for floor distribution relative to the passengers.

Now you have to look at the possibilities regarding passenger arrangements this is equivalent to a multinomial distribution with n = 5 trials and eight possibilities which is a generalized version of the binomial where you sample without replacement (in other words once the person gets off the elevator he is not put back in the sample), and the probability of each event will be 1/8 since all arrangements have the same probability.

Multinomial distribution - Wikipedia, the free encyclopedia

So basically you have this PDF and you need to find the probabilities that no two more passengers leave the same floor. This corresponds to having four of the people on board selecting a separate floor and the last person having the choice of selecting one from the remaining four, but you need to allow for all possibilities.

Now you can evaluate this probability in many ways. One way is to consider the complement of the situation and do one minus that and the other is to get the actual probability.

I'll get your response before we continue but there are many ways of evaluating things in mathematics.

Personally for this I might resort to using a computer by running a small program but I'm guessing you need to practice for some kind of entrance exam so this might not be appropriate for your situation.

Hey chiro,
Thanks for giving me hint to solve this problem. I know about multinomial distribution. If I calculate answer correctly, I definitely post it here.ok.

Hi,
The answer to this problem is 0.01465 assuming that all arrangements for discharging passengers have same probability.Is this correct?
But what would be answer if we don't assume same probabilities for all arrangements for discharging passengers?

Reply is expected from any member of this forum.