Probability-Has the owner made a smart decision?

Question

Problem: A high school locker room owner has a locker room with 1000 spaces. Each space costs Rs. 100 a day. He has already sold monthly permits to 1001 high school members (knowing that it is likely that not all the high school members would want to keep their bags there at the same time).

If someone with a permit arrives to keep their bag and there are no spaces, the owner will refund Rs. 200 for that day (penalty of Rs.100).

Parking on any given day of the month is independent of every other day.

Has he made the smart decision by selling an extra space?

What I am stuck at is the issue that there are no probabilities given in this question. As we are talking about high school members that means everyone will have lectures on different time of the day (6 working days-from monday to saturday). I think this question will not be requiring any numerical values to be solved. As expected profit is the part of this problem so E[random variable (RV] will be useful in solving this problem. Now If I am not wrong I think binomial RV will also play a major role in getting a general solution for this problem (because if we are interested in that 1 extra member then we will be interested in only two outcomes. 1. if that member will come 2. if that member will not come). That's all I know. I have no idea from where to start. I have no idea how to use CDF(Cumulative distribution function) to my advantage or how to even start this problem and get to the desired results. I am trying to find solution in terms of concepts related to RV not simply probability problems. I would appreciate if anyone can share their thoughts on this problem.

Answer 1

You want a model which means that I need to make a bunch of assumptions, given below.

• The owner can charge for the space to a person only if the person uses it on a particular day, so nothing can be done about "no shows".

• Each person has a uniform probability $p$ of needing a space on any day.

With the above assumptions, the "break even point" will occur if
P(all 1001 spaces are needed on a day) $ = 0.5$

For this, $p^{1001} = 0.5$ which yields $p = 0.999308....$

This is an extremely high value for $p$ which is most unlikely,
hence the owner has indeed made a smart decision !

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Added

If we assume $p=\frac{1000}{1001}$, which is still highly iunrealistic,

P(all 1001 spaces are needed) $\approx= 0.3677$, so the owner will lose , say $37$% of the time, and gain equivalent amounts $63$% of the time.

Answer 2

To begin with, a month has thirty days. You have called binomial distribution correctly - this is $X=B(30,p)$ that you're studying, wherein $p$ is the probability that all 1001 spaces are used at the same day. Calculate the latter, then the expected value of your variable times 100 is a good measure of whether the owner made a smart decision (I believe he did).

Answer 3

There is insufficient information here, but one can make straightforward assumptions and answer the question.

One assumes the distribution of users follows a Poisson distribution with a given mean $\mu$. Suppose you assume that $\mu = 1000$. That means some days there will be fewer than $1000$, some days more.

To get the expected return, you integrate (actually, it is a discrete distribution, so you merely sum) the values in your distribution (times 100 Rs) up to $x = 1000$ to get the expected income, and then sum your distribution from $1001$ (times 200 Rs) to infinity to find the expected loss. Subtract these two numbers to get the expected overall profit.

Answer 4

As true blue anil says you need a lot more information to make a model, then you can use the model to decide how many lockers to rent. It sounds to me like he rents the lockers on a monthly basis, or on some term longer than a day. Each day, some of the renters show up and use lockers. The simplest assumption is that each person shows up independently with probability $p$. If he has rented $1001$ so far and recorded how many people come each day he has a measure of $p$. We have enough people that the normal approximation is reasonable, so if he rents to $n$ people he would expect a mean of $np$ to show up with a standard deviation of $\sqrt{np(1-p)}$. If $k$ people show up his revenue for the day is $$r(n,k)=\begin {cases} 100n&k \le 1000\\100n-200(k-1000)& k\gt 1000 \end {cases}$$ We want to maximize the expected value of the revenue $E(r(n,k))$ by choosing $n$ given the probability distribution of $k$. We can convert the sum over $k$ to an integral in the continuous approximation and we get $$E(r(n,k))=100n-200\int_{1000}^\infty(k-1000)\frac 1{\sqrt {2\pi np(1-p)}}e^{-\frac{(k-np)^2}{2np(1-p)}}dk$$ Evaluate the integral, take the derivative with respect to $n$, set to $0 \ldots$

As a mathematical approach this is defensible if the model is right. Two problems are that there could be special events where $p$ gets much higher. Say there is a big soccer match in town. Maybe all you customers come in for the match and your refund loss is much higher. It could be that somebody watches the clients arrive and depart and discovers that most people show up by $9$ AM and nobody leaves before $11$. He signs up $50$ friends who rent lockers, then all show up between $9$ and $11$ hoping to collect refunds.