Hotel Rooms probability

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Jul 26th 2006, 01:47 PM
magicpunt

Hotel 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 PM
ThePerfectHacker

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 PM
magicpunt

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 PM
ThePerfectHacker

Quote:

Originally Posted by magicpunt

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

Thus probability of getting a hotel is .85 and you want to have $n$ such as probability at least one is .95 or more. If you have $n$ what is the probability that you do not get a room? Well, each one is .15 and there are $n$ thus the probability is OF GETTING AT LEAST ONE ROOM, is
$1-(.15)^n$ and you want that $P(n)\geq .95$ . Thus,
$1-(.15)^n\geq .95$
Thus,
$-(.15)^n\geq -.05$
Thus,
$(.15)^n\leq .05$
Make chart, for values,
$ \left\{ \begin{array}{cc} n&(.15)^n\\ 1&.1500\\ 2&.0225\\ 3&.0034 $
Note at $n=2$ we have already it less than .05

However, I am sure if this is what you mean.
Jul 26th 2006, 06:02 PM
magicpunt

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 AM
CaptainBlack

Quote:

Originally Posted by magicpunt

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).

Let P(N|M) be the probability that there are exactly N shows out of M bookings.

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 AM
JakeD

Quote:

Originally Posted by magicpunt

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).

If the manager books $n$ rooms, the number of rooms taken will have a binomial distribution with probability of success $p = .85$ and number of trials $n.$ You want to know what is the highest value of $n$ that leaves a .95 probability that the number of rooms taken is no more than 300. Since calculating that probability from the binomial distribution is messy and probably not intended by your instructor, you should use the normal approximation to the binomial distribution.

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

Quote:

Originally Posted by JakeD

$P(S_n \le 300) \ge .95 .$ Can you take it from here?

Continuity correction!

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

RonL
Jul 27th 2006, 02:11 AM
CaptainBlack

Quote:

Originally Posted by JakeD

Treat the number of rooms taken as a normal variable $S_n$ with mean $pn$ and standard deviation $\sqrt{p(1-p)/n}$

Standard deviation of a binomial RV with probability $p$ of
success on single trial and $n$ trials is:

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

RonL
Jul 27th 2006, 07:18 AM
JakeD

Quote:

Originally Posted by CaptainBlack

Standard deviation of a binomial RV with probability $p$ of
success on single trial and $n$ trials is:

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

RonL

Thank you, Captain! Confusing the standard deviation of $S_n /n,$ which is $\sqrt{p(1-p) /n},$ with the standard deviation of $S_n,$ which is $\sqrt{p(1-p)n}$ as you say, is a common mistake for me and it seems it's not the last time I'll make it. When working this kind of problem, I'll check whether the results are reasonable, and then have to go back and make that correction.

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

Quote:

Originally Posted by JakeD

As for the continuity correction, there ain't no stinking continuity correction in the Central Limit Theorem! ;)

Then how would you use the normal approximation to evaluate:

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