Probability (1 Viewer)

[untouchablecuz]

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?

(i) the first guy can choose n doors, so can the second guy and so on until the n-th guy. so, the number of ways is nⁿ

(ii) P(at least one door will not be chosen)=1-P(all doors chosen)

if all doors are chosen, the first guy will have n doors to choose from, the second guy will have n-1 doors to choose from and so on, until the n-th guy has one door left to choose

so, P(all doors chosen)=[n(n-1)...3.2.1]/nⁿ=n!/nⁿ

hence, P(at least one door will not be chosen)=1-n!/nⁿ

is my working for ii) correct?

 

[lolokay]

you are correct

 

[kwabon]

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?

(i) the first guy can choose n doors, so can the second guy and so on until the n-th guy. so, the number of ways is nⁿ

(ii) P(at least one door will not be chosen)=1-P(all doors chosen)

if all doors are chosen, the first guy will have n doors to choose from, the second guy will have n-1 doors to choose from and so on, until the n-th guy has one door left to choose

so, P(all doors chosen)=[n(n-1)...3.2.1]/nⁿ=n!/nⁿ

hence, P(at least one door will not be chosen)=1-n!/nⁿ

is my working for ii) correct?

thats a better way in fact.

(y)

 

[cutemouse]

I recall seeing this in a HSC paper. Which one was it from?

 

[untouchablecuz]

03

 

[shaon0]

You can use the above result of P=1-(n!/n^n) to prove that n^n>n! for n>1.

 

[cutemouse]

Umm, shouldn't the answer to part (ii) be:

Required Probability = [(1ⁿ + 2ⁿ + 3ⁿ + .. (n-1)ⁿ]/nⁿ

?

 

[untouchablecuz]

yep

thats coroneos' answer; he took a more direct approach

i was just checking to see if my solution was also correct

 

[untouchablecuz]

You can use the above result of P=1-(n!/n^n) to prove that n^n>n! for n>1.

is this what you had in mind?

 

[lolokay]

Umm, shouldn't the answer to part (ii) be:

Required Probability = [(1ⁿ + 2ⁿ + 3ⁿ + .. (n-1)ⁿ]/nⁿ

?

no, why?

(I think I might see where you're coming from - but there would be, for example, n ways to choose the (n-1) rooms to put the people in, and since this doesn't ensure that all the rooms will have at least one person in them, you will then be over counting many combinations)