HSC 2003 Q4c - Probability (1 Viewer)

[goobi]

A hall has n doors. Suppose that n people each choose any door at random to enter the hall.
(i) In how many ways can this be done? (ans= n^n)
(ii) What is the probability that at least one door will not be chosen by any of the people?


I have no problem with part 1, it's just part 2 that confuses me so much.
Apparently Coroneos and the Math Association provide two different answers for this question.

Coroneos:


Math Association


If you sub a number into n in both solutions, say 4, you will get two different answers i.e. 29/32 and 49/128 respectively.
So could someone please tell me which answer is the correct one or both?
Thanks

 

[someth1ng]

Math Association one looks correct.

 

[zhiying]

Okay I'm gonna try to do this without looking at either first and see which one matches...

whenever I see "at least" I always like to get 1-P(all) thing

P(all doors chosen) = n!/n^n, n! because since there's n doors and n people, so each person has to pick one door so first person has n choices, second has n-1, third n-2 and so on, n^n is the total possible ways from part a)

then P(at least one door not chosen) = 1 - P(all doors chosen) = 1-n!/n^n

Edit: yeah it fits the Math Association one, lol Coroneos tends to write overly long and complicated solutions from what I've seen in his 4u books -_-

 

[Carrotsticks]

The Coroneos solutions are incorrect.

They argue *paraphrased* "If 1 door is not chosen, then there are n-1 doors remaining, so each of the n people have n-1 choices and hence (n-1)^n".

This is incorrect because what guarantee is there that all of those n-1 doors are occupied? If at least one of those n-1 doors are *edit* NOT occupied, then we get the case when 2 doors are unoccupied in total. Furthermore, they didn't choose which doors to be unoccupied. So if they have 1 door unoccupied, they should have nC1 to choose the door to be unoccupied.