# probability. (1 Viewer)

#### underthesun

##### N1NJ4
Just curious.

I got i) n^n

and for the second, (n^n - n!)/n^n

is that what yous got?

i got i) n^n
ii) (n^n -n)/n^n

#### worlds greatest

##### Member
i got 1-(n!/n^n )
which is the same as underthesun

#### Constip8edSkunk

##### Joga Bonito
hmm i got 1- n^(1-n) which is same as tornado... me check now...

edit: changed mind... got it wrong... should be n! cuz there are n! arrangments for n peope choosing n doors... damn

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#### freaking_out

Originally posted by underthesun
Just curious.

I got i) n^n

and for the second, (n^n - n!)/n^n

is that what yous got?
i got the same, but i was scared, coz that looked simple-relative to the whole exam ofcourse.

#### enak

##### Limbo
Originally posted by Constip8edSkunk
should be n! cuz there are n! arrangments for n peope choosing n doors... damn
That's what I put down, I hope you're right

#### yeeha89

##### New Member
What? I am pretty sure that the first part was n^2, which is n square man.

#### Constip8edSkunk

##### Joga Bonito
i meant the n! in the numerator... like underthesun's answer instead of n... 1st part is n^n

#### Toodulu

##### werd!
hmm i think it's meant to be n^n
like 5 people and 5 doors.. each person has 5 choices.. so 5.5.5.5.5
ie. 5^5

N

#### ND

##### Guest
I misread the second part... i thought it said "one door will not be chosen", missed the "at least" part.

#### freaking_out

Originally posted by underthesun
lucky i read the question though....i was treating it as a counting problem, at the start.

#### hagun

##### New Member
Originally posted by worlds greatest
i got 1-(n!/n^n )
which is the same as underthesun
a. N^N
b. 1 - (N! / (N^N))

N

#### ND

##### Guest
Can someone tell me how they got the solution to ii)? This is what i would have done (had i read the question properly):

Consider a particular door, the probability that it will be walked through is 1/(n^n). There are n doors so probability is n/(n^n)=1/n^(n-1).

#### Constip8edSkunk

##### Joga Bonito
heres waht i thought

let x be the number of doors not opened
P(x>=1) = 1 - P(x=0)
P(x=0) = number of arrangements wif all doors chosen / total arrangments
=n!/n^n
.'. P(x>=1) = 1 - n!/n^n

i got n! mixed up with n

#### nerdd

##### Member
oh damn
i said n factorial for the first one and the second one was 1 - 1 on factorial to the n hahaha
so 1 - (1/n!)^n

#### walla

##### Satisfied Customer
hang on a sec
surely its quite simple
P(at least one door not chosen) = 1 - p(all doors chosen)
since there are n people and n doors,
the first person can choose n, the second person can choose (n-1) etc
then p(all doors chosen) = n!
then p(at least one door not chosen) = 1 - n!

#### walla

##### Satisfied Customer
oh god i really want to delete that post

#### ghoolz

##### New Member
I put:
i) n! because n ppl can choose the first door then (n-1)ppl can choose the second and (n-2) the third and so on which makes n!
ii) 1 - (1/n!)
which then = (n! - 1)/n!

not sure how i got the second one but i sounded good in the exam room

thinking about it now it could actually be n^n because the n ppl can all go thru the one door. <shit>

confusing at 12:45am maybe tomorrow (or later today) to do it.

#### hagun

##### New Member
Originally posted by Constip8edSkunk
heres waht i thought

let x be the number of doors not opened
P(x>=1) = 1 - P(x=0)
P(x=0) = number of arrangements wif all doors chosen / total arrangments
=n!/n^n
.'. P(x>=1) = 1 - n!/n^n

i got n! mixed up with n
In addition to that, I just wanna say that:
N! is the number of ways that each people through the same door
and from point i) N^N is the total number of ways...