probability hsc qns 2003 qns 4c) (1 Viewer)

[qqmore]

A hall has n doors. Suppose n people choose any door at random to enter the hall.

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

My approach was:
P(At least 1 door not chosen)
=1 - P(All doors chosen)

And P(All doors chosen) = n! / n^n (not sure if this is right)

So MY final answer:
P(At least 1 door not chosen) = 1 - (n! / n^n)


Can you please tell me what I'm doing wrong, its different from the ans in the solutions.

 

[u-borat]

that's the right answer?

 

[qqmore]

Well the solutions said:

[1^n + 2^n + 3^n + ...... + (n-1) ^n ] / n^n this is when they did it in cases.

But when i tested n = 2, the results differ from this solution and my own solution

 

[u-borat]

well ur solution is wrong cos ur answer is what is written in my past paper book answers.
and doing the questuion myself, im also getting your answer.

 

[qqmore]

Oh i see, the solutions are from coroneos publications ...... pretty dodgy. I assume your solutions are from maths association ?

 

[Iruka]

I agree. I have tried enumerating all the possibilities for the case n=3, and that
[1^n + 2^n + 3^n + ...... + (n-1) ^n ] / n^n formula just doesn't work.

 

[u-borat]

Oh i see, the solutions are from coroneos publications ...... pretty dodgy. I assume your solutions are from maths association ?

Yup mathematical assocation of nsw, if that means anything to you.

 

[Affinity]

Well the solutions said:

[1^n + 2^n + 3^n + ...... + (n-1) ^n ] / n^n this is when they did it in cases.

But when i tested n = 2, the results differ from this solution and my own solution

Dodgy solution.. here's probably what the author of that solution was thinking and what was wrong with that..


the author was trying to calculate P(atleast 1 not chosen) by
P(1 chosen) + P(2 chosen) + ... + P(n-1 chosen)

The precise and correct interpretation is
P(exactly 1 chosen) + P(exactly 2 chosen) + ... + P( exactly n-1 chosen)

What he/she did was:
P(first door chosen) + P(first 2 doors chosen) + ... + P(first n-1 doors chosen)

the problem with that is
1.) doesn't account for the permutation of doors
2.) overlapping between the events


if you work with the correct interpretation,
P(exactly 1 chosen) + P(exactly 2 chosen) + ... + P( exactly n-1 chosen)

you will get the correct answer.. but it's not that easy to do.. (unless you reduce it back to 1- P(all))

 

[kony]

yes Coroneos is wrong with this.

it has quite a few mistakes like these.