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I have this problem here and I'm very unsure of how to start this. I have an idea but I'm not sure where to go from a certain point. The problem says: A video rental store is analyzing a flat fee rental program it is planning to offer and the program allows a subscriber to rent up to $8$ movies a month for a flat monthly fee. Each rental costs the store $1.25$ dollars in processing and personnel fees. Let $p_n$ be the probability that a subscriber rents $n$ movies in one month. We have the following:

$1)$ $p_0=a$

$2)$ $p_n-p_{n+1}=c\gt 0$ where $c$ is some constant for $n=0,1,...,7$

$3)$ The probability of a subscriber renting fewer than $3$ movies in one month is $0.55$

The store would like to make a profit of $1$ dollar per rental on average. What should the monthly fee be for the program to achieve this profit?

My first instinct is to find the expected number of rentals per month and multiply it by $1.25$ to see how much it would cost. Then find the expected cost of this program and charge a dollar more than that. But I can't think of how to find the probability. Any help here?

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1 Answer 1

up vote 2 down vote accepted

Unless I am missing something so far, here's what I have:

$p_0 = a$, $p_k = a - k c$ when $k \in \{1, 2,\ldots,7\}$. We have no additional information on $p_8$. $p_k = 0 \; \forall \; k>8$.

The sum of all probabilities must equal one:

$$\sum_{n=0}^{\infty} p_n = 1$$

This means that

$$a + (a-c) + (a-2 c) + \ldots (a-7c) + p_8 = 8 a - 28 c + p_8 = 1 1$$

The other condition mandates that $p_0+p_1+p_2=0.55$. This implies that

$$3 a-3 c=0.55$$

We are required to find the expected number of $N$, the number of movies rented in a month:

$$E[N] = \sum_{n=0}^{\infty} n \, p_n $$

As presently constituted, the problem does not provide sufficient information to determine this expected value. We need to know something about $p_8$ to continue.

EDIT

The situation changes if the case $k=8$ is allowed in the 2nd condition above; that is, $p_8=a-8 c$. Then we have the system of equations:

$$\begin{align} & 9 a-36 c = 1 \\ & 3 a - 3 c = 0.55\\ \end{align}$$

from which we deduce that $a=\frac{37}{135}$ and $c=\frac{13}{540}$. The expected value calculation then takes the form

$$\begin{align} & E[N] = a \sum_{n=1}^{8} n - c \sum_{n=1}^{8} n^2 \\ & = a \frac{8 \cdot 9}{2} - c \frac{8 \cdot 9 \cdot 17}{6} \\ & = \frac{221}{45} \end{align}$$

So, at 1.25 per rental, the expected cost is then about 6.14, and at a desired profit of a dollar, the suggested fee would then be about 7.14 per month.

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Well I copied it exactly from the book I'm using and I thought the same thing. Would it make a difference if $2)$ said that $n$ could take the value $8$? –  TheHopefulActuary Jan 17 '13 at 5:11
    
Yes! In fact, I'll post a solution for that case. –  Ron Gordon Jan 17 '13 at 5:13
    
@Kyle: was this useful? –  Ron Gordon Jan 17 '13 at 6:13
    
Wouldn't you also have $a-5c=0.09$ because Probability greater than 3 is 0.45. So why doesn't that work out to one? –  yiyi Jan 17 '13 at 9:09
    
Ah, you did catch an error. It should be $9 a - 36 c = 1$. I'll fix. –  Ron Gordon Jan 17 '13 at 11:48
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