Question

A hall has 'n' doors. Suppose that 'n' people each choose any door at random to enter the hall. a) In how many ways can this be done? b) What is the probability that at least 1 door will not be chosen by any of the people?

Answer 1

a. `n^n`; b. `(n-1)^n/n^n`

Explanation:

Let's do part a. with numbers first.

We have a hall with 4 doors and 4 people are looking to enter it. For each person, they can choose any one of the four doors, and so the number of ways 4 people can enter is `4xx4xx4xx4=4^4=256`

And so if we have `n` doors and the same number of people, the number of ways we can have all the people enter the hall is `n^n`.

If we set things up so that no one enters at least 1 door, we can wall off that choice. Doing it with numbers, it means that each of the 4 people only have 3 doors to choose from, giving the number of ways they can now enter is `3xx3xx3xx3 = 3^4 = 81`.

The same holds true for `n` people - we'll have `n-1` doors to choose from, giving

`(n-1)^n`

The probability then of at least 1 door not being chosen is:

`(n-1)^n/n^n=((n-1)/n)^n`

To finish our numerical example using `n=4`, we end up with:

`(3/4)^4=81/256~=31.6%`