Question

It's 11:00pm, and in checking reservations, you discover that 4 guests have not yet arrived, and you just sold
your last room. At your hotel, the odds of each guest not showing, if they have not arrived by 11:00pm, is 65%. What are the odds
that you will need to "walk" a guest tonight?

Answer 1

Using the binomial distribution, it is found that there is a 0.985 = 98.5% probability that you will need to "walk" a guest tonight.

For each guest, there are only two possible outcomes, either they arrive, or they do not. The probability of a guest arriving is independent of any other guest, hence, the binomial distribution is used to solve this question.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• 4 guests have not yet arrived, hence n = 4.

• The probability that they show up is of 65%, hence p = 0.65.

If at least one shows up, you will need to "walk" a guest tonight, hence the probability is:

$P(X \geq 1) = 1 - P(X = 0)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.65)^{0}.(0.35)^{4} = 0.015$

Then:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.015 = 0.985$

0.985 = 98.5% probability that you will need to "walk" a guest tonight.

You can learn more about the binomial distribution at https://brainly.com/question/24863377