There are 300 hotel rooms at a hotel. 15% chance of cancellation for each room. What is the max # of rooms you can book with a 95% chance that you won't be overbooked.

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- Jul 26th 2006, 01:47 PMmagicpuntHotel Rooms probability
There are 300 hotel rooms at a hotel. 15% chance of cancellation for each room. What is the max # of rooms you can book with a 95% chance that you won't be overbooked.

- Jul 26th 2006, 02:24 PMThePerfectHacker
That do not make no sense to me.

Repeat question, from the way I understand it, the answer is impossible. - Jul 26th 2006, 04:14 PMmagicpunt
p(a reservation will be kept) = .85 and those who don't keep their reservation simply don't show up, thus taking up a room without paying for it

- Jul 26th 2006, 04:28 PMThePerfectHackerQuote:

Originally Posted by**magicpunt**

*at least one*is .95 or more. If you have $\displaystyle n$ what is the probability that you do not get a room? Well, each one is .15 and there are $\displaystyle n$ thus the probability is OF GETTING AT LEAST ONE ROOM, is

$\displaystyle 1-(.15)^n$ and you want that $\displaystyle P(n)\geq .95$. Thus,

$\displaystyle 1-(.15)^n\geq .95$

Thus,

$\displaystyle -(.15)^n\geq -.05$

Thus,

$\displaystyle (.15)^n\leq .05$

Make chart, for values,

$\displaystyle

\left\{

\begin{array}{cc}

n&(.15)^n\\

1&.1500\\

2&.0225\\

3&.0034

$

Note at $\displaystyle n=2$ we have already it less than .05

However, I am sure if this is what you mean. - Jul 26th 2006, 06:02 PMmagicpunt
I appreciate the help but let me rephrase.

You are the hotel manager. You have 300 hotel rooms in your hotel. There is an 85% chance that those who book a room will actually show up and a 15% chance that they will not (thus the room is "booked" but not used).

You want to book as many rooms as possible so that there is a 95% chance that all of those who book with you will have a room (a 95% confidence that you won't be overbooked). - Jul 27th 2006, 12:27 AMCaptainBlackQuote:

Originally Posted by**magicpunt**

Then the question is:

What is the largest M such that

P=P(0|M)+P(1|M) + P(2|M) + .. + P(300|M)>=0.95 ?

(that is: what is the maximum number of bookings M so that there is less

than a 5% chance of more than 300 shows?).

RonL - Jul 27th 2006, 12:33 AMJakeDQuote:

Originally Posted by**magicpunt**

Treat the number of rooms taken as a normal variable $\displaystyle S_n$ with mean $\displaystyle pn$ and standard deviation $\displaystyle \sqrt{p(1-p)/n} .$ Find the maximum $\displaystyle n$ with the probability $\displaystyle P(S_n \le 300) \ge .95 .$ Can you take it from here? - Jul 27th 2006, 02:07 AMCaptainBlackQuote:

Originally Posted by**JakeD**

$\displaystyle P(S_n \le 300.5) \ge .95 .$

RonL - Jul 27th 2006, 02:11 AMCaptainBlackQuote:

Originally Posted by**JakeD**

Standard deviation of a binomial RV with probability $\displaystyle p$ of

success on single trial and $\displaystyle n$ trials is:

$\displaystyle \sigma=\sqrt{n\ p(1-p)}$

RonL - Jul 27th 2006, 07:18 AMJakeDQuote:

Originally Posted by**CaptainBlack**

As for the continuity correction, there ain't no stinking continuity correction in the Central Limit Theorem! ;) - Jul 27th 2006, 09:06 AMCaptainBlackQuote:

Originally Posted by**JakeD**

B(721, 0.232, 10342) - the probability of 721 success in 10342 trials

each with a probability of a favourable outcome of 0.232? :)

RonL