Probability Question (1 Viewer)

[redfield]

A probability question from the 2003 HSC Exam. I have found some differing solutions from from popular textbooks, and am not sure which one is correct:

Q:4.(c) A hall has n doors. Suppose that n people choose any door at random to enter the hall.
(i) In how many ways can this be done?
(ii) What is the probability that at least one door will not be chosen by any of the people?

My Answer
(i) n^n ways
(ii) Consider conjugate event:
i.e. P(E) = 1 - P(people going through separate doors)
= 1 - (n!)/(n^n)

A solution given by one of my textbooks has for (ii) P = (1^n + 2^n + ... + (n-1)^n) / (n^n).

Is anyone able to verify my answer? Cheers in advance.

 

[lolokay]

i'm pretty sure your solution is correct

 

[LordPc]

(i) In how many ways can this be done?

each person has n choices, and there are n people, so yea n^n

(ii) What is the probability that at least one door will not be chosen by any of the people?

one door not unchosen
= All cases - no doors unchosen
= n^n - n!

P(no doors unchosen) is
first person can choose any door, so n choices.
2nd person has to pick a unique door so n-1 choices (since there are n doors and people, if there is a single double up, we will get 1 door not chosen.)
4rd person has to pick another door, so n-2 choices

so P(no doors unchosen) = n!

so there are n^n-n! cases, and for the probability, you divide by all cases,

so (n^n-n!)/n^n, which is what you got

so yea, u got it right